2023 Algebra 2

The in-class quizzes will have an individual and a group component. You will do a set of questions on your own (15-20 minutes). You will then do the same set of questions in a small group (15-20 minutes). I will collect the individual as well as the group answer sheet. Your mark will be a weighted average (80% individual + 20% group).

What is the point of the group component? Making correct arguments and catching wrong arguments is an essential part of mathematics. My hope is that you will spend the group component to convince each other of your answers, and in that process, hone your skills of argumentation.

Passing the mid-semester and the final exam (at least 50%) are hurdles to pass the course.

Let \(R\) be an integral domain that contains a field \(F\) as a sub-ring. Assume that \(R\) is finite dimensional when viewed as a vector space over \(F\). Prove that \(R\) is a field.

Let \(\alpha\) be a complex root of the polynomial \(x^{3}-3x+4\). Find the inverse of \(\alpha^2+\alpha+1\) in the form \(a + b \alpha + c \alpha^{2}\) with \(a, b, c \in \mathbf{Q}\).

\noindent The particular polynomial and element are not important. Your method should work in general.

Prove that the polynomial \(x^3+3x+3\) is irreducible over \(\mathbf{Q}[\sqrt[3]{2}]\).

For a positive integer \(n\), set \(\zeta_{n} = e^{2\pi i / n}\). Find all values of \(n\) such that \(\zeta_{n}\) has degree at most 3 over \(\mathbf{Q}\).

A field extension \(K/F\) is algebraic if every element of \(K\) is algebraic over \(F\). Let \(K/F\) and \(L/K\) be algebraic extensions. Prove that \(L/F\) is also an algebraic extension.

Let \(K = \mathbf{Q}(\alpha)\) where \(\alpha\) is a complex root of \(x^3-x-1\). Determine the irreducible polynomial for \(\gamma = 1 + \alpha^2\) over \(\mathbf{Q}\).

For this problem, first go through Section 5 (Construction with Ruler and Compass) to understand the proof of the following theorem (converse of what we did in class).

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\noindent Theorem: Suppose \(a \in \mathbf{R}\) be such that there exist field extensions \[ \mathbf{Q} = F_0 \subset F_1 \subset \dots \subset F_n\] with \(F_{i+1} = F_i[\sqrt d_i]\) for some \(d_i \in F_i\) and \(a \in F_n\). Then \(a\) is a constructible number.

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Decide whether the regular 9-gon is constructible by ruler and compass.

For this problem, first understand the proof of Proposition 15.3.3.

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\noindent Proposition: Let \(F\) be a field of characteristic not equal to 2. Then every quadratic extension \(K/F\) can be written as \(K = F(\delta)\) where \(\delta^2 \in F\).

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Let \(m, n \in \mathbf{Z}\). Determine when \(\mathbf{Q}(\sqrt m)\) and \(\mathbf{Q}(\sqrt n)\) are isomorphic.

Factor the polynomial \(x^{16}-x\) over \(\mathbf{F}_4\) and \(\mathbf{F}_8\).

Find a formula for the number of monic irreducible polynomials of degree \(n\) over the field \(\mathbf{F}_p\). If the general counting argument seems daunting, explain the case \(n = 60\) (for full credit).

Let \(f(x) = x^4-10x^2+1 \in \mathbf{Z}[x]\). (This is the minimal polynomial of \(\sqrt 2 + \sqrt 3\)). Prove that \(f(x)\) is irreducible in \(\mathbf{Z}[x]\) but reducible mod \(p\) for every prime \(p\).

We learned that every irreducible polynomial over a field \(F\) of characteristic 0 must have distinct roots in any extension field of \(F\). Find a counter-example to this statement in positive characteristic.

We learned that in characteristic 0, every finite extension \(F \subset K\) has a primitive element. Find a counter-example to this statement in positive characteristic.

Let \(F = \mathbf{Q}[\zeta_p]\). Find all automorphisms \(\phi \colon F \to F\). Describe the group \(\mathrm{Aut}(F)\).

Let \(F = \mathbf{Q}[\zeta_p]\) where \(p\) is an odd prime. Let \(K = F[2^{1/p}]\). Find all automorphisms \(\phi \colon K/F \to K/F\). Describe the group \(\mathrm{Aut}(K/F)\).

Let \(S \to T \to \mathbf{C}\) be a sequence of connected branched covers of degree 2.

1. Prove that there exists \(p \in \mathbf{C}\) such that the monodromy of a small loop around \(p\) is a cycle of type \((12)(34)\) or \((1234)\).
2. Construct a branched cover \(S \to \mathbf{C}\) of degree 4 such that (1) does not hold. You have constructed a degree 4 extension of \(\mathbf{C}(t)\) that does not decompose into a sequence of two degree 2 extensions.

The following may be useful.
Proposition: Let \(S \to \mathbf{C}\) be a branched cover where \(S\) is connected. Then there is a point \(p \in \mathbf{C}\) such that the monodromy of a small loop around \(p\) is a non-identity permutation.

Giving a map \(R[x]/f(x) \to K\) is equivalent to

1. giving a map \(R \to K\), and
2. specifying the image \(\alpha\) of \(x\), which must satisfy \(f(\alpha) = 0\).

Determine the degrees of the splitting fields of the following polynomials over \(\mathbf{Q}\):

1. \(x^3-2\)
2. \(x^4-1\)
3. \(x^4+1\)

Let \(K = \mathbf{Q}[\sqrt 2, \sqrt 3, \sqrt 5]\). Determine \(\deg K / \mathbf{Q}\), prove that \(K/ \mathbf{Q}\) is a Galois extension, and determine its Galois group.

Let \(p\) be an odd prime number and \(K = \mathbf{Q}[\zeta_{p}]\). Prove that \(K\) contains a unique degree 2 extension of \(\mathbf{Q}\).

Find quartic polynomials in \(\mathbf{Q}[x]\) whose Galois group is isomorphic to:

1. The Dihedral group \(D_4\) (of order 8)
2. The cyclic group \(C_4\)

Remark: The general version of the above problem is a longstanding open problem called the Inverse Galois Problem: given a finite group \(G\), does there always exist a polynomial in \(\mathbf{Q}[x]\) with Galois group isomorphic to \(G\)?

Let \(\delta \in \mathbf{Q}\) be such that \(\mathbf{Q}[\sqrt \delta]\) is the unique degree 2 extension of \(\mathbf{Q}\) contained in \(\mathbf{Q}[\zeta_{p}]\). For \(p = 7\), find \(\delta\).

This is a continuation of the last problem. You should now know the subfield \(\mathbf{Q}[\sqrt \delta] \subset \mathbf{Q}[\zeta_p]\) for \(p = 3, 5, 7\). Based on this data, make a conjecture for an arbitrary odd prime \(p\). (If you need more data, work out the case of \(p = 11\).) Then try to prove the conjecture.

Some nested square roots can be de-coupled to a linear combination of simple square roots. For example, we have \[ \sqrt {5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3.\] But some cannot be. Prove that \(\alpha = \sqrt{1 + \sqrt 3}\) cannot be written as a sum \[ \sqrt{a_{1}} + \cdots + \sqrt{a_{n}}, \quad a_i \in \mathbf{Q}.\]

Hint. Compare the Galois group of the minimal polynomial of \(\alpha\) over \(\mathbf{Q}\) and the Galois group of \(\mathbf{Q}[\sqrt{a_1}, \dots, \sqrt{a_n}] / \mathbf{Q}.\)

Let \(\alpha \in \mathbf{C}\) be a nested square root. Let \(G\) be the Galois group of the minimal polynomial of \(\alpha\) over \(\mathbf{Q}\). Prove that the order of \(G\) is a power of \(2\).

Caution. Make sure that the extension you are considering is Galois!

Prove the converse to the problem before: if \(\alpha \in \mathbf{C}\) is such that its minimal polynomial over \(\mathbf{Q}\) has Galois group whose order is a power of 2, then \(\alpha\) is a nested square root. As an application, show that if \(p\) is a prime number of the form \(2^n+1\), then \(\zeta_p\) is a nested square root.

With this, we have completed a proof of the following.
Theorem. For a prime number \(p\), the regular \(p\)-gon is constructible if and only if \(p\) has the form \(2^n+1\).

In this problem, you may use the following fact from group theory without proof.
Theorem. Let \(p\) be a prime and \(G\) a group of order \(p^{n}\) for \(n \geq 1\). Then \(G\) contains a normal subgroup of index \(p\).

Determine the Galois group of the polynomial \(x^6+3\) over the base fields

1. \(F = \mathbf{Q}\)
2. \(F = \mathbf{Q}[\zeta_3]\).

Find a polynomial of degree \(7\) over \(\mathbf{Q}\) whose Galois group is \(S_7\).

Hint. Take inspiration from the construction in Artin for degree 5.

The simplest kinds of numbers are whole numbers, or elements of the ring \(\mathbf{Z}\). The next simplest are quotients of integers, or elements of the ring \(\mathbf{Q}\). Most of our “number rings” will be extensions of these.

The simplest kinds of functions are polynomial functions, or elements of the ring \(\mathbf{R}[x]\). (You can replace \(\mathbf{R}\) by others, like \(\mathbf{C}\) or \(\mathbf{Q}\)). The next simplest are the rational functions, or elements of \(\mathbf{R}(x)\). Most of our “function rings” will be extensions of these.

There are other extremely important rings related to the integers. These are the rings \(\mathbf{Z}/n \mathbf{Z}\). We will mostly focus on the case of \(n = p\) a prime number, so that the ring above is a field. We will meet extensions of these, which are strictly speaking neither “function rings” or “number rings”, and not strictly speaking are both.

Fix a number \(\alpha \in \mathbf{C}\). We denote by \(\mathbf{Z}[\alpha] \subset \mathbf{C}\) the smallest sub-ring of \(\mathbf{C}\) containing \(\mathbf{Z}\) and \(\alpha\). In other words, see for yourself that \[ \mathbf{Z}[\alpha] = \{a_{0} + a_1 \alpha + \dots + a_{n}\alpha^{n} \mid a_i \in \mathbf{Z}\}.\] More generally, given \(\alpha_{1}, \dots, \alpha_{m} \in \mathbf{C}\), we denote by \(\mathbf{Z}[\alpha_{1}, \dots, \alpha_{m}] \subset \mathbf{C}\) the smallest subring of \(\mathbf{C}\) containing \(\mathbf{Z}\) and \(\alpha_1, \dots, \alpha_{m}\). We can write elements of \(\mathbf{Z}[\alpha_{1},\dots, \alpha_{m}]\) just as before. More precisely, we have the following observation.

Let \(R \subset S\) be a sub-ring. Given \(\alpha_{1}, \dots, \alpha_{m}\), we define \(R[\alpha_{1}, \dots, \alpha_{m}]\) as the smallest subring of \(S\) that contains \(R\) and each of \(\alpha_{1}, \dots, \alpha_{m}\). As before, it is the image of the homomorphism \[ R[x_{1}, \dots, x_{n}] \to S\] that sends \(R \to S\) by the given inclusion and sends \(x_{i}\) to \(\alpha_{i}\).

If \(\alpha\) is algebraic, then \(F[\alpha]\) is a field.

Take \(\alpha = \sqrt[3]{2}\), for example, which is algebraic over \(\mathbf{Q}\). Then Proposition 8.2.2 says that \(\mathbf{Q}[\alpha]\) is a field. How will you explicitly find the multiplicative inverse of its elements? For example, what is the inverse of \(\sqrt[3]{2}\)? Of \(1 + \sqrt[3]{2}\)?

Let \(\alpha, \beta \in K\) be algebraic over \(F\). There is an isomorphism \(F[\alpha] \to F[\beta]\) extending the identity of \(F\) that sends \(\alpha \mapsto \beta\) if and only if \(\alpha\) and \(\beta\) have the same minimal polynomial.

We have the following dichotomy.

Let \(L = \mathbf{Q}[i, \sqrt 2]\). Then \(\deg (L / \mathbf{Q}) = 4\).

Let \(F \subset L\) and \(L \subset K\) be finite extensions. Then \(F \subset K\) is also finite and \[ \deg(F/K) = \deg(L/F) \cdot \deg (L/K).\]

If \(\alpha, \beta\) are algebraic over \(F\) then \(\alpha\beta\) and \(\alpha+\beta\) are algebraic over \(F\).

So we know that \(\sqrt 2 + \sqrt 3\) is algebraic. But what is its minimal polynomial? The method of proof does not really give a way to find out. One way to find a polynomial is by taking repeated powers and looking for a linear relation, but we still need tools to prove that the polynomial we found is irreducible.

The degree of \(\mathbf{Q}[\sqrt[3]2, i] / \mathbf{Q}\) is 6.

If \(F \subset L\) is an extension of degree \(M\), then the degree of all sub-extensions must divide \(M\). In particular, the degree of every \(\alpha \in L\) must divide \(M\).

For example, an extension of degree \(2^{n}\) cannot contain an element of degree 3.

• A non-trivial factorisation of \(p \in R\) is an expression \(p = ab\) where neither \(a\) nor \(b\) is a unit.
• We say that \(p\) is irreducible if it has no non-trivial factorisation.
• In general, a factorisation of \(f\) into irreducibles is a factorisation \(f = p_1 \cdots p_n\) where each \(p_{i}\) is irreducible.
• We say that two factorisations \(f = p_1 \cdots p_{n}\) and \(f = q_1 \cdots q_{m}\) are equivalent if \(m = n\) and after re-numbering, we have for all \(i)\) an equality \(f_i = g_i \times u_i\) where \(u_i\) is a unit.
• We say that \(R\) has unique factorisation or is a Unique Factorisation Domain if every element of \(R\) has a factorisation into irreducibles and this factorisation is unique up to equivalence.

• \(\mathbf{Z}\) is a UFD.

Let \(F\) be a field. Then \(F[x]\) is a UFD.

Let us come back to the main question: how do we show that a given polynomial is irreducible?

Let us relate factorisation over \(\mathbf{Q}[x]\) and \(\mathbf{Z}[x]\). Consider the factorisation \[ 3x+3 = 3 \cdot (x+1).\] This is a non-trivial factorisation in \(\mathbf{Z}[x]\) but trivial one in \(\mathbf{Q}[x]\). This is basically the only difference in the theory.

To get rid of factorisations as above, we introduce the following notion. We say that \(f(x) \in \mathbf{Z}[x]\) is a primitive polynomial if no prime \(p \in \mathbf{Z}\) divides \(f(x)\). Equivalently, for every prime \(p\), the image of \(f(x)\) in \(\mathbf{Z}/p \mathbf{Z}[x]\) is non-zero. Equivalently, we cannot extract a non-unit constant factor. By convention, we also require the leading coefficient of \(f(x)\) to be positive (this is less important; if we don’t do this, we have to amend most of the following statements by adding “up to sign” or “up to a unit in \(\mathbf{Z}\)”).

If \(f(x)\in \mathbf{Z}[x]\) is not primitive, we can simply extract out all primes \(p\) that divide it. In other words, we can write \[ f(x) = c g(x)\] where \(c \in \mathbf{Z}\) and \(g(x)\) is primitive.

More generally, if \(f(x) \in \mathbf{Q}[x]\), then we can take a common denominator of all coefficients and write \(f(x) = 1/N \cdot h(x)\) where \(h(x) \in \mathbf{Z}[x]\). We can then extract a constant factor out of \(h(x)\) if any and get \[ f(x) = c g(x) \] where \(c \in \mathbf{Q}\) and \(g(x) \in \mathbf{Z}[x]\) is primitive. It is easy to check that the expression above is unique.

Slogan: For primitive polynomials, factorisation in \(\mathbf{Q}[x]\) and \(\mathbf{Z}[x]\) behave in the same way.

We will make this precise. First, we need an easy lemma.

1. Given two points, construct their midpoint.
2. Given three points \(A, B, C\), construct the angle bisector.
3. Given three points \(A, B, C\), construct the unique circle passing through \(A, B, C\).
4. Given two points \(A, B\), divide the segment \(AB\) in 79 equal parts (or any other number).
5. …

To bring algebra into the picture, we introduce coordinates. Let a set of constructed points \(S\) be given. Suppose \(F \subset \mathbf{R}\) is a field that contains all the coordinates of \(S\). The key idea is to explore in what ways \(F\) needs to be enlarged when we construct new points.

Let \(P\) be a point constructed using the ruler and compass from a given set \(S\). Assume that the coordinates of \(S\) lie in a field \(F\). Then there exist extensions \[ F_{0} = F \subset F_1 \subset \dots \subset F_{n}\] of the form \(F_{i+1} = F_i[\sqrt a_i]\) for some \(a_i \in F_i\) and such that the coordinates of \(P\) lie in \(F_{n}\).

In particular, the degree of the extension of \(F\) generated by the coordinates of \(P\) is a power of 2.

If the coordinates of \(P\) generate a transcendental extension of \(F\) or an extension whose degree is not a power of 2, then \(P\) cannot be constructed from \(S\) using ruler and compass.

We start with the points \((0,0)\) and \((0,1)\). Then the point \((\cos 20, \sin 20)\) cannot be constructed. In particular, the \(60\)-degree angle cannot be trisected, and hence there cannot exist a procedure that trisects a given angle.

The field of complex numbers is algebraically closed.

Let \(\overline Q \subset \mathbf{C}\) be the set of numbers that are algebraic over \(\mathbf{Q}\). Then \(\overline Q\) is algebraically closed.

The only irreducible polynomials over \(\mathbf{R}\) are linear and quadratics with negative discriminant.

Funnily enough, there is no purely algebraic proof of the Fundamental Theorem of Algebra! This is somewhat expected because the construction of \(\mathbf{C}\) goes via the construction of \(\mathbf{R}\), which is quite non-algebraic. So at some point in the proof, some analysis or topology comes in. I consider this is a spectacular example of the unity of mathematics: how different fields of mathematics help each other!

Proposition Let \(F\) be a field and let \(p(x) \in F[x]\) be a non-constant polynomial. Then there exists a finite extension \(F \subset K\) such that \(p(x)\) has a root in \(K\). Proof Let \(f(x)\) be an irreducible factor of \(p(x)\). Take \(K = F[x]/f(x)\). Then \(\alpha = [x]\) is a root of \(p(x)\).

Suggestion: How do you represent elements of \(K\)? If \(f(x)\) has degree \(n\), then the elements can be identified with polynomials in \(F[x]\) of degree up to \(n-1\), with addition and multiplication done modulo \(f(x)\). Psychologically, I find it useful to rename \(x\) to a letter from the beginning of the alphabet, like \(a\) or \(\alpha\) so that the elements of \(K\) feel more like numbers than polynomials. This makes it less confusing if we have to make further extensions of \(K\), freeing up \(x\) for polynomials.

Proposition Let \(F\) be a field and let \(p(x) \in F[x]\) be a non-constant polynomial. Then there exists a finite extension \(F \subset K\) such that \(p(x)\) splits into linear factors in \(K[x]\). Proof Adjoin a root. Factor it out. Adjoin a root of the remaining polynomial. Rinse, repeat.

Definition We say that \(K/F\) is a splitting field of \(p(x)\) if \(p(x)\) splits into linear factors in \(K[x]\) and \(K\) is generated by the roots of \(p\).

The second condition ensures that \(p(x)\) does not factor completely over a subfield of \(K\).

Recall that the characteristic of a ring \(R\) is the smallest integer \(n\) such that \(n = 0\) in \(R\). Equivalently, it is the generator of the kernel of the map \(\mathbf{Z} \to R\). If \(R\) is a domain, then the generator must be a prime number. In particular, if \(R = F\) is a finite field, then the kernel must be \((p)\) for some \(p\). Then we have an injection \(\mathbf{F}_p \to F\).

Proposition Every finite field \(F\) admits a unique injection \(\mathbf{F}_p \to F\) where \(p\) is the characteristic of \(F\).

Let \(F\) be a finite field. Suppose \(F / \mathbf{F}_p\) is an extension of degree \(r\). Then \(F\) is an \(r\)-dimensional \(\mathbf{F}_p\) vector space. In particular, it has \(q = p^r\) elements.

Let \(R\) be a ring of characteristic \(p\). Then the map \(x \mapsto x^p\) is a ring homomorphism, called the Frobenius map. In particular, every finite field admits a Frobenius \(\phi \colon F \to F\). Since \(F\) is a field, \(\phi\) is injective, and since \(F\) is finite, it is also surjective.

The multiplicative group \(F^{\times}\) has order \(q-1\), so for every \(x \in F\) we have \(x^{q-1} = 1\). Multiplying by \(x\), we get \(x^q = x\) for every \(x \in F\). In terms of the Frobenius, the above means that \[ \phi^r = \operatorname{id}.\]

Indeed, let \(\alpha\) be a generator of \(F^{\times}\). Then \(F = \mathbf{F}_p[\alpha]\), which is isomorphic to \(\mathbf{F}_p[x]/f(x)\) where \(f(x)\) is the minimal polynomial for \(\alpha\).

For every \(q\), does there exist a finite field with \(q\) elements? Equivalently, for every \(r\), do we have an irreducible polynomial in \(\mathbf{F}_p[x]\) of degree \(r\)?

The answer is Yes! Let us construct \(F\) with \(q\) elements.

We know that the elements of such \(F\) have to be roots of \(x^q-x = 0\). So, to construct \(F\), we just formally adjoin the roots! More precisely, we know that there is a finite extension \(K / \mathbf{F}_p\) such that \(x^q-x\) splits into linear factors in \(K\).

Proposition The polynomial \(x^q-x\) has distinct roots in \(K\).

To prove this we have to take a slight digression and understand multiple roots.

Let \(F\) be any field. We define the derivative of a polynomial formally. That is, if \(f(x) = \sum a_i x^{i}\) then \(f'(x) = \sum i a_i x^{i-1}.\) With this definition, the sum/product/chain rules continue to hold!

Proposition If \(\alpha \in F\) is a multiple root of \(f(x)\), then \((x-\alpha)\) divides \(f(x)\) and \(f'(x)\). Corollary If \(f\) has a multiple root then \(\gcd(f(x), f'(x))\) is non-constant.

Consider the polynomial \(x^q - x\). Its derivative is \(1\)! So it cannot possible have multiple roots! As a result, it has \(q\) distinct roots in \(K\).

Proposition The set of roots of \(x^q-x\) forms a subfield of \(K\).

We have now constructed (!) a field with \(q\) elements.

Thanks to what we have proved, we know that there exists an irreducible polynomial of every given degree. We find one and set \(F = \mathbf{F}_p[x]/f(x)\).

Note that there are many choices for \(f\) and hence we get many possible \(F\). But… wait for it….

Let \(F\) and \(L\) be finite fields of size \(q = p^r\). Then there is an isomorphism \(F \cong L\). In fact, there are exactly \(r\) isomorphisms.

Let \(F\) be a field of size \(q = p^{r}\). If \(F\) contains a field of size \(p^{s}\) then \(s\) divides \(r\). Conversely, if \(s\) divides \(r\), then \(F\) contains a unique subfield of size \(p^{s}\).

Before we begin, let us collect a few useful facts that we already know. Fix fields \(F \subset K\).

Proposition Suppose we have \(f(x), g(x) \in K[x]\) that actually lie in \(F[x]\). Then their (monic) gcd \(\gcd(f,g)\) also lies in \(F[x]\). Proof There are many ways to see this. The most direct is to observe that the gcd can be computed by Euclid’s algorithm, which will never leave \(F[x]\) if it starts with two polynomials in \(F[x]\). Another way is to use that \(\gcd(f,g)\) can be written as \(f(x) a(x) + g(x) b(x)\) for \(a(x), b(x) \in F[x]\).

The next observation is about repeated roots and derivatives. Proposition: If \(\gcd(f, f') = 1\), then \(f\) has no repeated roots in \(K\).

Note that if \(f(x) \in F[x]\), then \(f'(x) \in F[x]\) and so the gcd will also live in \(F[x]\).

Corollary: If \(f\) is irreducible in \(F[x]\) and \(f' \neq 0\), then \(f\) has no repeated roots in \(K\). Proof: If \(f\) is irreducible in \(F[x]\), and \(f' \neq 0\), then \(f' \in F[x]\) has lower degree than \(f\). So \(\gcd(f,f')\), which lies in \(F[x]\), must be 1.

Corollary: If \(f\) is irreducible and \(F\) is of characteristic 0, then \(f\) has no repeated roots in \(K\). Proof: We may assume \(f\) is non-constant. Then \(f' != 0\) is automatic in characteristic 0 (but not in characteristic \(p\)!).

To be frank, the statement of the primitive element theorem is more important than the proof. Even within the proof, the statement of the claim below is more important than the rest of the details. It gives an explicit construction of a primitive element.

It suffices to prove that \(K = \mathbf{F}[\alpha, \beta]\) has a primitive element (we then induct). Let \(f,g \in F[x]\) be min polys of \(\alpha\) and \(\beta\). Choose an extension \(L / K\) in which both \(f\) and \(g\) split completely. (If you want to keep things concrete, imagine \(F = \mathbf{Q}\) and \(L = \mathbf{C}\) or \(\overline{\mathbf{Q}}\).) Then \(f\) and \(g\) split into distinct factors. Let \(\alpha_i \in L\) be the roots of \(f\) and \(\beta_j \in L\) the roots of \(g\). Say \(\alpha = \alpha_1\) and \(\beta = \beta_1\).

We prove that for all but finitely many \(\lambda \in F\), the element \(\gamma = \alpha + \lambda \beta\) is a primitive element. More precisely, choose \(c \in F\) such that the elements \[ \gamma_{ij} = \alpha_i + c \beta_j\] are all distinct. (This excludes only finitely many \(c\), and since our field \(F\) is necessarily infinite, leaves infinitely many choices.)

Claim With \(c\) chosen as above, \(\gamma\) is a primitive element for \(K = F[\alpha,\beta]\).

We now prove the claim. Let \(M = F[\gamma] \subset K\). We want to prove that \(M = K\). It is enough to prove that \(\alpha \in M\); because then \(\beta = (\gamma - \alpha)/c \in M\) and so \(M = K\).

To prove that \(\alpha \in M\), we prove that the minimal polynomial \(h(x)\) of \(\alpha\) over \(M\) has degree 1. To see this, observe that we can write down two polynomials in \(M[x]\) satisfied by \(\alpha\), namely \(f(x)\) and also \(g((\gamma - x)/c)\). Then \(h(x)\) divides both of them. Both \(f(x)\) and \(g(\gamma - cx)\) split completely over \(L\). But see that they only have one common root: namely \(\alpha = \alpha_1\). Indeed, the roots of \(f(x)\) are \(\alpha_i\) and the roots of \(g((\gamma - x)/c)\) are \(\gamma - c \beta_{j}\). We have \[ \alpha_i = c\beta_j - \gamma\] if and only if \[ \alpha_i + c\beta_j = \gamma,\] which happens only for one \(i\) and \(j\), namely \(i = 1\) and \(j = 1\). (This is because we chose \(c\) very carefully.) So \(f(x)\) and \(g((\gamma - x)/c)\) cannot have a common factor of degree > 1 in \(L\). It follows that \(\deg h = 1\).

Here is an example of a field extension without a primitive element. It has to be in characteristic \(p\), and it has to be over an infinite field. So take \(F = \mathbf{F}_p(x,y)\) and let \(K = F[u,v]/(u^p-x, v^p-y)\). This is the field obtained by adjoining the \(p\)-th roots of \(x\) and \(y\).

You can do it in sequence, if you want. That is, let \(K_1 = F[u]/(u^p-x)\)—and verify somehow that \(u^p-x\) is irreducible in \(F[u]\). And then \(K = K_1[v]/(v^p-y)\)—and again verify somehow that \(v^p-y\) is irreducible in \(K[v]\).

Then \(\deg(K/F) = p^2\). But note that the \(p\)-th power of any element of \(K\) lies in \(F\). So no element of \(K\) has degree \(p^2\) over \(F\). As a result, there is no primitive element.

To every irreducible \(f(t,x) \in \mathbf{C}[t,x]\) of positive \(x\)-degree, we can associate a closed subset \(S(f) \subset \mathbf{C}^2\). It is defined by \[ S(f) = \{(t,x) \in \mathbf{C}^2 \mid f(t,x) = 0\}.\] We can view elements of \(K\) as functions on \(S(f)\) (minus a finite set of points). We have a map \(S(f) \to \mathbf{C}\) given by \((t,x) \mapsto t\).

Here are some examples of the suraces obtained in this way. The picture is a projection to \(\mathbf{R}^3\) of the actual surface in \(\mathbf{R}^4 = \mathbf{C}^2\). The missing fourth coordinate is indicated by the colour. The map to \(\mathbf{C}\) is the projection to the horizontal plane (down).

1. \(f(t,x) = x^2 - t\):

1. \(f(t,x) = x^3 - t\):

Project: Write a computer program to generate these pictures from an \(f\).

For most values of \(t\), we expect \(f_t(x)\) to have \(n\) distinct roots. This is indeed what happens. In fact, more is true. Let \(\pi \colon S(f) \to \mathbf{C}\) be the projection.

Theorem: There exists a finite set \(B \subset \mathbf{C}^2\) such that \[ \pi^{-1}(\mathbf{C} -B) \to \mathbf{C}-B\] is a connected \(n\)-sheeted covering space.

The finite set is not unique—we can always enlarge it and the statement still holds.

Definition An \(n\)-sheeted branched cover of \(\mathbf{C}\) is a connected covering space of \(\mathbf{C} - B\) for some finite set \(B\).

The set \(B\) is unimportant. We can always enlarge it and we treat the resulting covering space as representing the same branched cover.

Consider two branched covers represented by covering spaces \(S_1 \to \mathbf{C}^2-B\) and \(S_2 \to \mathbf{C}^2-B\). A map of branched covers is a continuous map \(f \colon S_1 \to S_2\) that commutes with the projections to \(\mathbf{C}^2-B\). That is, it maps a point of \(S_1\) lying over \(t\) to a point of \(S_2\) lying over the same \(t\).

The construction \(K_f \mapsto S(f)\) gives us a procedure \[ \text{Field extension of } \mathbf{C}(t) \to \text{Branched cover of } \mathbf{C}.\] A map of field extensions yields a map of corresponding branched covers (in the other direction)!

To see how, suppose we have a map \(\phi \colon K_g \to K_f\). Suppose \(\phi(x) = h(t,x)\). Then \[ (t,x) \mapsto (t, h(t,x))\] gives a map from \(S(f) \to S(g)\).

Example Take \(f(t,x) = x^6-t(t+1)^{2}\) and \(g(t,x) = x^2 - t\). Then we have a map \(K_g \to K_f\) given by \(x \mapsto x^3/(t+1)\). We have an induced map \(S(f) \to S(g)\) given by \[ (t,x) \mapsto (t,x^3/(t+1)).\] See that if \((t,x)\) lies on \(S(f)\) then the image indeed lies on \(S(g)\).

A remarkable theorem is that the correspondence above is an equivalence.

Theorem (Riemann Existence Theorem): The procedure above is an equivalence (“equivalence of categories”). This means that this procedure is (1) a bijection between isomorphism classes of extensions \(K_f / \mathbf{C}(t)\) and branched covers \(S(f) \to \mathbf{C}\) (2) a bijection between maps of field extensions and maps of the corresponding branched covers.

In short, studying field extensions of \(\mathbf{C}(t)\) is equivalent to studying branched covers of \(\mathbf{C}\)!

Remark: Given a branched cover, it is not at all clear how to find the polynomial \(f\). Simalarly, given a map of branched covers it is not at all clear why it should be induced by an algebraic map. This is the hard part in the theorem.

Covering spaces are characterised by their monodromy. This is a simple, beautiful, and ubiquitous geometric concept.th) Let \(\pi \colon X \to Y\) be an \(n\)-sheeted covering space (for us \(Y\) will be \(\mathbf{C}^2 - B\)). Fix a basepoint \(y \in Y\), and label its pre-image in \(X\) by \(1,2,\dots,n\). Take a walk \(\gamma\) in \(Y\) starting at \(y\) and coming back to \(y\). After we choose a starting point in \(X\) from \(1,\dots, n\), there is a unique way to lift this walk up to \(X\). This is because \(X \to Y\) is a covering space, so the pre-image of a small neighborhood in \(Y\) consists of \(n\) disjoint copies of the same neighborhood. We just have to continue in whatever copy we are in!

The following picture (\(y = p^{*}\)) shows this in an example.

The lift may not return to the starting point, but it must return to one of \(1, \dots, n\). We thus get a permutation \(p_{\gamma}\) of \(1,\dots, n\). The map \(\gamma \mapsto p_{\gamma}\) is called the monodromy.

Theorem \(p_{\gamma}\) depends only on the homotopy class of \(\gamma\).

That is, a continuous perturbation of \(\gamma\) does not change \(p_{\gamma}\).

Let us consider \(Y = \mathbf{C}^2 \setminus B\). Choose a system of curves \(\gamma_{i}\) as shown below (the crosses are points of \(B\)):

It turns out that up to homotopy, any curve in \(\mathbf{C}-B\) is a concatenation of \(\gamma_i\)’s or their reverses. So, to specify the monodromy, it suffices to specify the permutations \(p_i\) associated to each \(\gamma_{i}\).

Theorem Given any permutations \(p_1, \dots, p_{b}\), there is a covering space of \(\mathbf{C}-B\) (unique up to isomorphism) whose monodromy is \(\gamma_{i} \mapsto p_i\).

Proof (sketch) We choose half rays starting at each cross and extending to infinity. We cut the plane open along these rays, and stack \(n\) copies of these cut planes on top of each other. We label the copies \(1,\dots, n\). The monodromy tells us exactly how to glue these sheets together along the cut edges (see picture).

(Caution: The result may not be connected. To ensure it is, we must be able to go from any \(i\) to any \(j\) by a repeated application of \(p_1, \dots, p_{b}\) and their inverses. But this is a minor point.)

For simplicity, let us take our coefficients to be in \(\mathbf{Q}\). Suppose \(\alpha_1, \dots, \alpha_n\) are the roots. Let \(F = \mathbf{Q}[\alpha_1, \dots, \alpha_n] \subset \mathbf{C}\). This \(F\) is called the splitting field of the polynomial. We would like to describe the field extension \[ \mathbf{Q} \subset \mathbf{Q}[\alpha_1, \dots, \alpha_n].\] In particular, we want to know if we can arrive at this extension by a sequence of easier extensions like \[ K \subset K[a^{1/n}].\] Galois theory was developed to answer exactly questions of this form.

Galois theory gives us tools to answer questions of the following kind: Given a field extension \(F \subset L\), can we decompose it into smaller extensions of a particular type? For example, is there any non-trivial decomposition \(F \subset K \subset L\)? That is, is there any intermediate field \(K\)?

The main theorem of Galois theory tells us about all possible intermediate fields!

Let us consider \[ \mathbf{Q} \subset F = \mathbf{Q}[\alpha_1, \dots, \alpha_{n}].\] The main insight of Galois theory is that the nature of this extension is controlled by symmetries among the roots \(\alpha_{1}, \dots, \alpha_n\) (to be made precise).

Key Observation: Let \(\phi\) be an automorphism of the field extension \(F / \mathbf{Q}\), that is, an automorphism \(\phi \colon F \to F\) that restricts to the identity on \(\mathbf{Q}\). Then \(\phi\) permutes \(\alpha_1, \dots, \alpha_n\).

Key insight: Not all permutations of \(\alpha_1, \dots, \alpha_n\) arise from automorphism \(\phi\). But if we understand which ones do, then we understand the nature of the field extension.

Let \(\operatorname{Aut}(F / \mathbf{Q})\) be the group of automorphisms of \(F/\mathbf{Q}\). By the observation above, we have a homomorphism \[ \operatorname{Aut}(F / \mathbf{Q} ) \to S_{n}\] Since the \(\alpha_i\) generate \(F\), the only automorphism that fixes all \(\alpha_{i}\) is the identity. That is, the homomorphism above has trivial kernel, so it is injective. The Galois group of the polynomial \(x^n+a_{n-1}x^{n-1}+ \dots +a_{0}\) is the image of the homomorphism above. By the first isomorphism theorem, it is an isomorphic copy of \(\operatorname{Aut}(F / \mathbf{Q} )\).

Theorem. (Main theorem of Galois theory): There is a bijective correspondence between fields sandwiched in \(\mathbf{Q} \subset F\) and subgroups of the Galois group.

We will what the bijection is in due course. Here, let us explain how to go from a subgroup to a subfield: \[ H \leadsto F^H = \{x \in F \mid h(x) = x \text{ for all } h \in H\}.\]

Let us take \(n = 2\), and consider \(p = u_{1}^3 + u_{2}^3\). Then \(p\) is symmetric, and indeed we have \[ p = (u_{1}+u_{2})^3 - 3u_{1}u_{2}(u_{1}+u_{2}) = s_{1}^3-3s_{1}s_{2}.\]

Any symmetric polynomial expression in the roots can be written as a polynomial expression in the coefficients.

A particularly important symmetric expression of the roots is the discriminant \[ D = \prod_{i

Let us keep a running example, say \(n = 3\) and \(p = u_{1}^3+u_{2}^3+u_{3}^{3}\). We will prove existence. Uniqueness is less important (and follows from the same proof).

We use a double induction: on \(n\) and on the degree of \(p\). There is not much to prove for \(n = 1\). Assume the statement for \(n-1\) and for all polynomials of degree smaller than \(p\). Consider \(p(u_{1}, \dots, u_{n-1},0)\). This is symmetric in \(u_{1}, \dots, u_{n-1}\), and by the inductive hypotheses, can be written as a polynomial in \(s_{1}(u_{1}, \dots, u_{n-1}), \dots, s_{n-1}(u_{1}, \dots, u_{n-1})\), say \[ p(u_{1}, \dots, u_{n-1}, 0) = h(s_{i}(u_{1}, \dots, u_{n-1})).\]

In our case, \[p(u_{1},u_{2}, 0) = u_{1}^{3}+u_{2}^3 = s_{1}^{3}-3s_{1}s_{2}.\]

We now take the expression for \(p(u_{1}, \dots, u_{n-1}, 0)\) in terms of \(s_{i}(u_{1}, \dots, u_{n-1})\), and replace \(s_{i}(u_{1}, \dots, u_{n-1})\) by \(s_{i}(u_{1}, \dots, u_{n-1}, u_{n})\). That is, we consider \[ q = h(s_{i}(u_{1}, \dots, u_{n})).\]

In our case, \[ q = (u_{1}+u_{2}+u_{3})^3 - 3(u_{1}+u_{2}+u_{3})(u_{1}u_{2} +u_{2}u_{3} + u_{1}u_{3}) = u_{1}^{3}+u_{2}^{3}+u_{3}^{3} - 3u_{1}u_{2}u_{3}. \]

We now consider the difference \[ p(u_{1}, \dots, u_{n}) - q(u_{1}, \dots, u_{n}).\] There is no reason that this is zero (we are lucky if it is!). But by construction, it is zero if we substitute \(u_{n} = 0\). That is, the difference is divisible by \(u_{n}\). But see that the difference is symmetric! So it must be divisible by \(u_{1}, \dots, u_{n}\), and hence by their product! So \[ p-q = u_{1}\cdots u_{n} r(u_{1}, \dots, u_{n}),\] where \(r\) is also symmetric. But \(r\) has smaller degree, so it can be expressed in terms of the elementary symmetric polynomials. As a result, so can \(p\).

In our case, \(r = 3\), so there’s nothing to do.

Take a rational polynomial \(f(x)\). It splits into linear factors over \(\mathbf{C}\), say \[ f(x) = (x-\alpha_{1}) \cdots (x-\alpha_{n}).\] Then \(K = \mathbf{Q}[\alpha_1,\dots,\alpha_n]\) is a splitting field of \(f(x)\) over \(\mathbf{Q}\).

The extension \(\mathbf{F}_{p^{n}}/ \mathbf{F}_p\) is a splitting field of

• any irreducible polynomial in \(\mathbf{F}_p[x]\) of degree \(n\)
• the polynomial \(x^{p^n}-x\)

Consider \(K = \mathbf{Q}[2^{1/3}] / \mathbf{Q}\). It is not a splitting field of \(x^3-2\). Is it a splitting field of some other polynomial? (Spoiler: No, see the next theorem.)

Theorem Let \(K/F\) be a splitting field of \(f(x)\). Suppose \(g(x) \in F[x]\) is an irreducible polynomial that has a root in \(K\). Then \(g(x)\) splits into linear factors in \(K[x]\).

Consider \(K = \mathbf{Q}[2^{1/3}] / \mathbf{Q}\). Then \(x^3-2\) has a root in \(K\) but does not split completely in \(K[x]\). So \(K / \mathbf{Q}\) cannot be a splitting field of any polynomial.

Let \(K_1/F\) and \(K_2/F\) be splitting fields of \(f(x) \in F[x]\). Then there exists an isomorphism \[ K_1/F \to K_2/F.\]

Lemma Let \(K_1/F\) and \(K_2/F\) be finite extensions. There exists a finite extension \(K/F\) such that contains isomorphic copies of \(K_1/F\) and \(K_2/F\). That is, there exist subfields \(K_{1}' \subset K\) and \(K_2' \subset K\) with isomorphisms \(K_{i}/F \cong K_{i}'/F\).

Let \(K = \mathbf{Q}[\zeta]\) where \(\zeta = e^{2\pi i/5}\) and \(F = \mathbf{Q}\). Then \(K/F\) has degree 4 and it is the splitting field of \[ x^{5} - 1.\] It is also the splitting field of \[ x^4+x^3+x^2+x+1,\] which is in fact the minimal polynomial of \(\zeta\).

Let \(G = \operatorname{Aut}(K/F)\). An element of \(G\) is determined by where it sends \(\zeta\). Furthermore, it must send \(\zeta\) to a root of its minimal polynomial. So there are only 4 possible elements of \(G\) corresponding to \[ \zeta \mapsto \zeta^{i}\] for \(i = 1,2,3,4\). But since \(K/F\) is Galois, we know that it has 4 automorphisms. As a result, all 4 choices must give valid automorphisms.

The map \[ i \mapsto \text{The automorphism that sends } \zeta \to \zeta^{i}\] gives an isomorphism of groups \[ \mathbf{Z}/ 5 \mathbf{Z}^{\times} \to G.\] The group on the LHS is cyclic of order 4, generated by \(2\) for example, so the Galois group is also cyclic of order 4. Henceforth, we identify these two groups.

The diagram of subgroups of \(G\) is:

So, by the main theorem, the diagram of intermediate fields is:

Observe that each step in this tower is a quadratic extension, and hence is obtained by adjoining a square root. As a result, \(\zeta\) can be expressed in terms of two iterated square roots.

Let us dig deeper and find an explicitl such expression for \(\zeta\). As a first step, let us identify the extension \(L\). By the main theorem, \(L\) is the fixed field of the subgroup \(\{1,4\}\). The element \(1\) corresponds to the identity automorphism, which fixes everything. So \(L\) is the fixed field of the automorphism corresponding to \(4\), which sends \(\zeta \to \zeta^{4}\). This automorphism sends

In the last line, we used \[ \zeta^4 = -1 - \zeta - \zeta^2 -\zeta^3.\] We see that for this element to be fixed, we must have \[ a_1 = 0 \quad\text{and} \quad a_2=a_3. \] So \(L\) consists of elements of the form \[ a_{0} + a_2(\zeta^2+\zeta^3),\] where \(a_0,a_2 \in \mathbf{Q}\). Set \(\alpha = \zeta^2+\zeta^3\). Then \[ L = \mathbf{Q}[\alpha].\]

We still haven’t found out which square root we need to adjoin to get \(L\). To do so, we find the minimal polynomial for \(\alpha\) over \(\mathbf{Q}\). We can do it as before—find a relation between powers of \(\alpha\). But using automorphisms, we can do better (as we did using Frobenius). The roots of the minimal polynomial are (by definition) the conjugates of \(\alpha\) over \(\mathbf{Q}\). We have the following.

Proposition Let \(K/F\) be a Galois extension with Galois group \(G\). The \(F\)-conjugates of \(\alpha \in K\) are the precisely the elements in the \(G\)-orbit of \(\alpha\). In other words, if \(\{\alpha_1, \dots, \alpha_{n}\}\) is the \(G\)-orbit of \(\alpha\), then \[ f(x) = (x-\alpha_1) \cdots (x-\alpha_{n})\] is the minimal polynomial for \(\alpha\) over \(F\).

(Proof sketch: We see that \(f(x)\) is fixed by \(G\) so lives in \(F[x]\). So the minimal polynomial must divide \(f(x)\). On the other hand, every \(\alpha_i\) must be a root of the minimal polynomial for \(\alpha\). So \(f(x)\) is the minimal polynomial. This is Theorem 16.5.2(a) in Artin).

Coming back to our situation, the conjugates of \(\alpha\) are \[ \alpha = \zeta^2 + \zeta^3 \text{ and } \beta = \zeta + \zeta^4.\] So the minimal polynomial for \(\alpha\) is

(In the calculation, we use \(1+\dots+\zeta^4 = 0\)). So, we get \[ \alpha = \frac{-1 \pm \sqrt{5}}{2}.\] In particular, \[ L = \mathbf{Q}[\sqrt 5].\]

Finally, to write \(\zeta\), we find its minimal polynomial over \(L\). Let us again use the conjugates. Since \(\operatorname{Aut}(K/L) = \{1,4\}\), the \(L\)-conjugates of \(\zeta\) are \[ \zeta \text{ and } \zeta^4.\] So the minimal polynomial of \(\zeta\) over \(L\) is: \[ (x-\zeta)(x-\zeta^4) = x^2 + (\alpha+1) x + 1.\] (In the calculation, we use \(1+\dots+\zeta^4 = 0\)). So we arrive at an explicit equation for \(\zeta\) using iterated square roots:

Let \(F = \mathbf{Q}\) and \(K = \mathbf{Q}[\zeta_3, 2^{1/3}]\). We know that \(K/F\) is Galois of degree 6 and the Galois group \(G\) is isomorphic to \(S_3\) via its action on the three roots of \(x^3-2\).

Let us number the roots \(\alpha_1 = 2^{1/3}\), \(\alpha_2 = 2^{1/3}\zeta_3\), and \(\alpha_3 = 2^{1/3}\zeta_3^2\). Consider the subgroup \(H = \{e, (23)\} \subset S_3\). Then the corresponding \(L\) is \(L = \mathbf{Q}[2^{1/3}]\). The extension \(L/F\) is not Galois and indeed the subgroup \(H \subset G\) is not normal.

On the other hand, consider the subgroup \(H = \{e,(123),(132)\}\). The corresponding \(L\) is \(L = \mathbf{Q}[\zeta_3]\). The extension \(L/F\) is Galois and indeed the subgroup \(H \subset G\) is normal.

The key observation is the following. Claim. Suppose \(L/F\) is Galois and \(g \in \operatorname{Aut}(K/F)\). Then \(g\) sends all elements of \(L\) to \(L\). Proof. Take \(\alpha \in L\). Let \(f(x) \in F[x]\) be the minimal polynomial of \(\alpha\). Then \(g(\alpha) \in K\) must be a root of \(f(x)\). Since \(L/F\) is Galois, it is a splitting field, so it contains all roots of \(f(x)\).

We now prove one direction: if \(L/F\) is Galois, then \(H = \operatorname{Aut}(K/L)\) is a normal subgroup. Let \(h \in H\) and \(g \in G\). We need to show that \(g^{-1} h g \in H\). That is, we need to show that \(g^{-1} h g\) fixes \(L\). Take \(\alpha \in L\). By the claim, we have \(g(\alpha) \in L\); so \(h\) fixes \(g(\alpha).\) That is, we have \[ h g (\alpha) = g(\alpha). \] But this is equivalent to \[ g^{-1}h g(\alpha) = \alpha.\] Since \(\alpha \in L\) was arbitrary, we have proved that \(g^{-1} h g\) fixes \(L\).

Let us prove the other direction. Recall that \(L = K^{H}\) is the fixed field of \(H\). Suppose \(H \subset G\) is normal. Again, the key observation is the converse to the above claim. Claim. Suppose an intermediate field \(L\) has the property that for all \(g \in \operatorname{Aut}(K/F)\), the element \(g\) sends \(L\) to itself. Then \(L/F\) is Galois. Proof. We prove that \(L/F\) is a splitting field. Take \(\alpha \in L\) primitive over \(F\) with minimal polynomial \(f(x)\). Since \(K/F\) is Galois, \(f(x)\) splits completely over \(K\). It is enough to prove that all the roots of \(f(x)\) actually lie in \(L\). But the roots of \(f(x)\) are of the form \(g(\alpha)\) for \(g \in G\). Since \(g\) sends \(L\) to itself, we see that \(g(\alpha) \in L\).

We now prove the other direction. Take \(g \in G\) and \(\alpha \in L\). We need to show that \(g(\alpha) \in L\). But \(L\) is the fixed field of \(H\); so we need to show that \(g(\alpha)\) is fixed by all \(h \in H\). That is, we need to show that \[ h g (\alpha) = g(\alpha),\] or equivalently \[ g^{-1}h g(\alpha) = \alpha.\] But note that \(g^{-1}h g \in H\) because \(H \subset G\) is normal, so the last equation holds because all elements of \(H\) fix \(\alpha \in L\).

Let \(L\) be an intermediate field of \(K/F\) such that \(L/F\) is itself Galois. Let \(H = \operatorname{Aut}(K/L)\) be the corresponding (normal) subgroup of \(H\).

Proposition. We have a surjective homomorphism \[ \operatorname{Aut}(K/F) \to \operatorname{Aut}(L/F)\] whose kernel is \(H = \operatorname{Aut}(K/L).\)

Proof. The most non-obvious point is the existence of the homomorphism. Given \(g \in \operatorname{Aut}(K/F)\), we would like to restrict it to obtain an element of \(\operatorname{Aut}(L/F)\). For an arbitrary \(L\), there is no reason to believe that \(g\) sends \(L\) to itself, so the restriction does not make sense. But if \(L/F\) is normal, then it does (by one of the claims we just proved). So we indeed get a restriction map \[ \operatorname{Aut}(K/F) \to \operatorname{Aut}(L/F).\] Its kernel consists of the automorphisms of \(K\) that fix \(L\), that is, \(\operatorname{Aut}(K/L)\). It is not immediately clear why it is surjective, but if we look at the sizes of the three groups, we see that it must be surjective.

Let us begin by observing that a computer (running sage, for example) knows how to compute \(G\) given \(f(x)\).

R = QQ['x']
f = R(x^4+2*x^2+3)
f.galois_group().list()

[(), (1,3)(2,4), (1,4,3,2), (1,2,3,4), (2,4), (1,3), (1,4)(2,3), (1,2)(3,4)]

Here is another example.

R = QQ['x']
f = R(x^4+5*x+5)
f.galois_group().list()

[(), (1,2,3,4), (1,3)(2,4), (1,4,3,2)]

Here is another example that looks just like the previous example.

R = QQ['x']
f = R(x^4+3*x+3)
f.galois_group().list()

[(), (1,3)(2,4), (1,4,3,2), (1,2,3,4), (2,4), (1,3), (1,4)(2,3), (1,2)(3,4)]

but the answer is quite different!

[(), (1,3)(2,4), (1,4,3,2), (1,2,3,4), (2,4), (1,3), (1,4)(2,3), (1,2)(3,4)]

Let me emphasise that we (at least, I) can compute Galois groups by hand only for easy polynomials. Easy means either low degree or of a special form.

The basic principle of how we compute Galois groups by hand is not too far from the general procedure above.

The basic principle. Suppose we have an algebraic relation among the roots, that is, a polynomial in \(F[x_1, \dots, x_n]\) that is satisfied when substitute \(\alpha_{i}\) for \(x_i\). Then any element of the Galois group must preserve that relation.

We start with a general result.

Proposition. Let \(F\) be a field of characteristic 0 and let \(f(x) \in F[x]\) be irreducible of degree \(n\). Let \(K/F\) be the splitting field of \(f(x)\) and \(\alpha_1, \dots, \alpha_n \in K\) the roots of \(f(x)\). Then the roots form one orbit under the Galois group \(G\).

Proof (sketch). If the roots formed two orbits, then taking the product of \((x-\alpha)\) over one orbit and over the other orbit gives two polynomials fixed by \(G\) whose product is \(f(x)\). Polynomials fixed by \(G\) have coefficients in \(F\), so this gives a non-trivial factorisation of \(f(x)\). For three or more orbits, we do the same thing; we get even more factors.

A subgroup \(G \subset S_n\) under which \(\{1,\dots, n\}\) forms only one orbit are called transitive. Up to re-numbering there are only a handful of transitive subgroups of \(S_4\):

1. \(S_4\)
2. \(A_4\)
3. \(D_4\)
4. \(C_4 = \langle (1234) \rangle\)
5. \(D_{2} = V = \{(), (12)(34), (14)(23), (13)(24)\}\).

The key principle in finding the Galois group by hand is the following.

Principle Find a polynomial relation \(R\) in the roots \(\alpha_i\) with coefficients in \(F\).

For example, \(\alpha_2^2 = 3\alpha_2\) is a relation. A relation \(R\), by itself, gives no information. But if \(R\) is a relation, and \(p\) is a permutation, we can look at the relation \(R'\) obtained by permuting the roots according to \(p\).

Idea. If the relation \(R'\) does not hold then \(p\) cannot be in the Galois group.

If \(R\) is a symmetric relation, then \(R' = R\), so we get no information. The key is to find non-symmetric relations.

The trouble is that it is difficult to compute values of non-symmetric expressions. But we can compute them indirectly! It’s best to see this in an example.

Proposition Let \(G\) be a solvable group. Then every subgroup of \(G\) is solvable and every quotient of \(G\) is solvable.

Proposition. Let \(G\) be a group and \(N \subset G\) a normal subgroup. If \(N\) and \(G/N\) are both solvable, then \(G\) is solvable.

Corollary. Products of solvable groups are solvable. Finite abelian groups are solvable.

Let \(f, g \in F[x]\) and let \(K / F\) be a splitting field of \(fg\). Inside \(K\), the ring generated by the roots of \(f\) is a splitting field of \(f\) and similarly for \(g\). Call these \(K_f\) and \(K_g\), respectively. Note that both \(K_f/F\) and \(K_g/F\) are Galois. We have restriction maps \[ \operatorname{Aut}(K/F) \to \operatorname{Aut}(K_f/F),\] and similarly for \(g\). By putting the two together, we have a map \[ \operatorname{Aut}(K/F) \to \operatorname{Aut}(K_f/F) \times \operatorname{Aut}(K_g/F).\] Proposition The map above is injective. Proof. An automorphism of a splitting field is determined by where it sends the roots. The roots of \(fg\) are the union of the roots of \(f\) and roots of \(g\). So if an automorphism fixes the roots of \(f\) and \(g\), it must fix the roots of \(fg\).

The proposition says that the Galois group of \(fg\) is a subgroup of the product of the Galois groups of \(f\) and the Galois group of \(g\).

Let us recall the theorem again. Theorem. Let \(f(x) \in F[x]\) be irreducible with splitting field \(K/F\). The following are equivalent.

1. A root of \(f(x)\) is a nested radical over \(F\).
2. All roots are
3. \(G = \operatorname{Aut}(K/F)\) is solvable.

We will prove the theorem under the assumption that \(F\) contains all \(p\)-th roots of unity for all \(p \leq \deg K\). Let us also assume that \(F\) is a subfield of \(\mathbf{C}\). We can then make all our constructions within \(\mathbf{C}\) and it also makes it easier to talk about \(p\)-th roots. Neither of these are serious restrictions.

2 implies 1 is clear.

3 implies 2 is also fairly easy. Suppose \(G\) is solvable. Consider a filtration \[ 1 = G_{0} \subset G_{1} \subset \cdots \subset G_{n} = G\] where \(G_{i} \subset G_{i+1}\) is normal and the quotient is cyclic of prime order. (The primes are necessarily less that the order of \(G\).) Consider the corresponding chain of fields \[ K = K_{n} \supset K_{n-1} \supset \cdots \supset K_{0} = F.\] Then \(K_{i} \subset K_{i+1}\) is a Galois extension with Galois group cyclic of prime order. By Kummer’s theorem, \(K_{i}\) is obtained by adjoining a \(p\)-th root. So all elements of \(K\), in particular all the roots of \(f(x)\), are nested radicals.

Let us prove that 1 implies 3. 1 says that we have a tower of fields \[ F = F_{0} \subset F_{1} \subset \cdots \subset F_{n}\] obtained by adjoining \(p\)-th roots and \(F_n\) contains a root of \(f(x)\). The problem is that \(F_{n}/F\) need not be Galois. To remedy this, we take the given tower and convert it to a tower of fields which are Galois over \(F\).

We construct fields \(K_i\) such that

1. \(F_i \subset K_i\)
2. \(K_i / F\) is Galois
3. \(\operatorname{Aut}(K_{i+1}/K_{i})\) is a subgroup of a product of cyclic groups of order \(p\).

To get started, we set \(K_{0} = F\) and \(K_1 = F_1\). Let us think about constructing \(K_{2}\). We know that \(F_2\) is obtained from \(F_{1}\) by adjoining a \(p\)-th root, say a \(p\)-th root of \(a\).