# 07-401/Class Notes for April 11

## Today's Agenda

• Today's the deadline for the prize problem from 07-401/Homework Assignment 7!
• Reminder of the Fundamental Theorem of Galois Theory.
• Proof of the insolubility of the quintic assuming the Fundamental Theorem.
• Proofs of the easy parts of the Fundamental Theorem.
• A short discussion of the final and and the time leading to it.
• Course Evaluation Forms and a Post-Mortem discussion in the spirit of 0506-1300/Post Mortem and of 06-240/Classnotes For Thursday December 7.
• With luck, early dismissal!

## The Final Exam

As announced by the powers above, our final exam will take place on the evening of Tuesday April 24 between 7PM and 10PM, at New College Residence (NR) room 25.

The exam will be similar in style to the Term Test (also see On the Term Test). The material is everything covered in class. Everything in the test will be taken from our text book, and there will be two types of questions (or maybe sometimes the two types will be mixed within a single question):

• You may be asked to prove a theorem proven in class. The reason we prove theorems in class is that these proofs are valuable. Therefore I expect you to know them.
• You may be asked to solve exercises from the relevant chapters of the book, or minor variations thereof. These may be questions that were assigned as homework, but also, these may be questions that were not assigned before.

Office Hours. I (Dror) will hold extended office hours before the final, on Monday April 23 11AM-1PM and on the exam date, Wednesday April 24 10AM-12PM. You will be able to pick up all your graded assignments then and also on my last "normal" office hour, on Wednesday April 18 10:30AM-11:30AM. All office hours will be held at or near my office, Bahen 6178

Preparing for the Test. Read, reread and rereread everything and solve lots of exercises from the book.

My (Dror's) system when I was an undergrad was to prepare a 4-6 page 100-200 item list of points covered in class. I'd only summarize each point with one sentence, without giving any details and without trying to be precise, much like the list that I prepared for the class of February 7 (see On the Term Test). I would then go over my list again and again and again, crossing out every item for which I was sure I could complete all the details and supply all the proofs. I would only stop when there was nothing left to cross out.

Good Luck!

## The Fundamental Theorem of Galois Theory

Note. An updated version of this section is at 08-401/The Fundamental Theorem.

It seems we will not have time to prove the Fundamental Theorem of Galois Theory in full. Thus this note is about what we will be missing. The statement appearing here, which is a weak version of the full theorem, is taken from Gallian's book and is meant to match our discussion in class. The proof is taken from Hungerford's book, except modified to fit our notations and conventions and simplified as per our weakened requirements.

Here and everywhere below our base field $F$ will be a field of characteristic 0.

### Statement

Theorem. Let $E$ be a splitting field over $F$. Then there is a bijective correspondence between the set $\{K:E/K/F\}$ of intermediate field extensions $K$ lying between $F$ and $E$ and the set $\{H:H<\operatorname{Gal}(E/F)\}$ of subgroups $H$ of the Galois group $\operatorname{Gal}(E/F)$ of the original extension $E/F$:

$\{K:E/K/F\}\quad\leftrightarrow\quad\{H:H<\operatorname{Gal}(E/F)\}$.

The bijection is given by mapping every intermediate extension $K$ to the subgroup $\operatorname{Gal}(E/K)$ of elements in $\operatorname{Gal}(E/F)$ that preserve $K$,

$\Phi:\quad K\mapsto\operatorname{Gal}(E/K)$,

and reversely, by mapping every subgroup $H$ of $\operatorname{Gal}(E/F)$ to its fixed field $E_H$:

$\Psi:\quad H\mapsto E_H$.

This correspondence has the following further properties:

1. It is inclusion-reversing: if $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$ and if $K_1\subset K_2$ then $\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_2)$.
2. It is degree/index respecting: $[E:K]=|\operatorname{Gal}(E/K)|$ and $[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.
3. Splitting fields correspond to normal subgroups: If $K$ in $E/K/F$ is the splitting field of a polynomial in $F[x]$ then $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.

### Lemmas

The two lemmas below belong to earlier chapters but we skipped them in class.

#### The Primitive Element Theorem

The celebrated "Primitive Element Theorem" is just a lemma for us:

Lemma 1. Let $a$ and $b$ be algebraic elements of some extension $E$ of $F$. Then there exists a single element $c$ of $E$ so that $F(a,b)=F(c)$. (And so by induction, every finite extension of $E$ is "simple", meaning, is generated by a single element, called "a primitive element" for that extension).

Proof. See the proof of Theorem 21.6 on page 375 of Gallian's book. $\Box$

#### Splitting Fields are Good at Splitting

Lemma 2. (Compare with Hungerford's Theorem 10.15 on page 355). If $E$ is a splitting field of some polynomial $f$ over $F$ and some irreducible polynomial $p\in F[x]$ has a root $v$ in $E$, then $p$ splits in $E$.

Proof. Let $L$ be a splitting field of $p$ over $E$. We need to show that if $w$ is a root of $p$ in $L$, then $w\in E$ (so all the roots of $p$ are in $E$ and hence $p$ splits in $E$). Consider the two extensions

$E=E(v)/F(v)$ and $E(w)/F(w)$.

The "smaller fields" $F(v)$ and $F(w)$ in these two extensions are isomorphic as they both arise by adding a root of the same irreducible polynomial ($p$) to the base field $F$. The "larger fields" $E=E(v)$ and $E(w)$ in these two extensions are both the splitting fields of the same polynomial ($f$) over the respective "small fields", as $E/F$ is a splitting extension for $f$ and we can use the sub-lemma below. Thus by the uniqueness of splitting extensions, the isomorphism between $F(v)$ and $F(w)$ extends to an isomorphism between $E=E(v)$ and $E(w)$, and in particular these two fields are isomorphic and so $[E:F]=[E(v):F]=[E(w):F]$. Since all the degrees involved are finite it follows from the last equality and from $[E(w):F]=[E(w):E][E:F]$ that $[E(w):E]=1$ and therefore $E(w)=E$. Therefore $w\in E$. $\Box$

Sub-lemma. If $E/F$ is a splitting extension of some polynomial $f\in F[x]$ and $z$ is an element of some larger extension $L$ of $E$, then $E(z)/F(z)$ is also a splitting extension of $f$.

Proof. Let $u_1,\ldots,u_n$ be all the roots of $f$ in $E$. Then they remain roots of $f$ in $E(z)$, and since $f$ completely splits already in $E$, these are all the roots of $f$ in $E(z)$. So

$E(z)=F(u_1,\ldots,u_n)(z)=F(z)(u_1,\ldots,u_n)$,

and $E(z)$ is obtained by adding all the roots of $f$ to $F(z)$. $\Box$

### Proof of The Fundamental Theorem

#### The Bijection

Proof of $\Psi\circ\Phi=I$. More precisely, we need to show that if $K$ is an intermediate field between $E$ and $F$, then $E_{\operatorname{Gal}(E/K)}=K$. The inclusion $E_{\operatorname{Gal}(E/K)}\supset K$ is easy, so we turn to prove the other inclusion. Let $v\in E-K$ be an element of $E$ which is not in $K$. We need to show that there is some automorphism $\phi\in\operatorname{Gal}(E/K)$ for which $\phi(v)\neq v$; if such a $\phi$ exists it follows that $v\not\in E_{\operatorname{Gal}(E/K)}$ and this implies the other inclusion. So let $p$ be the minimal polynomial of $v$ over $K$. It is not of degree 1; if it was, we'd have that $v\in K$ contradicting the choice of $v$. By lemma 2 and using the fact that $E$ is a splitting extension, we know that $p$ splits in $E$, so $E$ contains all the roots of $p$. Over a field of characteristic 0 irreducible polynomials cannot have multiple roots and hence $p$ must have at least one other root; call it $w$. Since $v$ and $w$ have the same minimal polynomial over $K$, we know that $K(v)$ and $K(w)$ are isomorphic; furthermore, there is an isomorphism $\phi_0:K(v)\to K(w)$ so that $\phi_0|_K=I$ yet $\phi_0(v)=w$. But $E$ is a splitting field of some polynomial $f$ over $F$ and hence also over $K(v)$ and over $K(w)$. By the uniqueness of splitting fields, the isomorphism $\phi_0$ can be extended to an isomorphism $\phi:E\to E$; i.e., to an automorphism of $E$. but then $\phi|_K=\phi_0|_K=I$ so $\phi\in\operatorname{Gal}(E/K)$, yet $\phi(v)=w\neq v$, as required. $\Box$

Proof of $\Phi\circ\Psi=I$. More precisely we need to show that if $H<\operatorname{Gal}(E/F)$ is a subgroup of the Galois group of $E$ over $F$, then $H=\operatorname{Gal}(E/E_H)$. The inclusion $H<\operatorname{Gal}(E/E_H)$ is easy. Note that $H$ is finite since we've proven previously that Galois groups of finite extensions are finite and hence $\operatorname{Gal}(E/F)$ is finite. We will prove the following sequence of inequalities:

$|H|\leq|\operatorname{Gal}(E/E_H)|\leq [E:E_H]\leq |H|$

This sequence and the finiteness of $|H|$ imply that these quantities are all equal and since $H<\operatorname{Gal}(E/E_H)$ it follows that $H=\operatorname{Gal}(E/E_H)$ as required.

The first inequality above follows immediately from the inclusion $H<\operatorname{Gal}(E/E_H)$.

By the Primitive Element Theorem (Lemma 1) we know that there is some element $u\in E$ so that $E=E_H(u)$. Let $p$ be the minimal polynomial of $u$ over $E_H$. Distinct elements of $\operatorname{Gal}(E/E_H)$ map $u$ to distinct roots of $p$, but $p$ has exactly $\deg p$ roots. Hence $|\operatorname{Gal}(E/E_H)|\leq\deg p=[E:E_H]$, proving the second inequality above.

Let $\sigma_1,\ldots,\sigma_n$ be an enumeration of all the elements of $H$, let $u_i:=\sigma_iu$ (with $u$ as above), and let $f$ be the polynomial

$f=\prod_{i=1}^n(x-u_i)$.

Clearly, $f\in E[x]$. Furthermore, if $\tau\in H$, then left multiplication by $\tau$ permutes the $\sigma_i$'s (this is always true in groups), and hence the sequence $(\tau u_i=\tau\sigma u_i)_{i=1}^n$ is a permutation of the sequence $(u_i)_{i=1}^n$, hence

$\tau f=\prod_{i=1}^n(x-\tau u_i)=\prod_{i=1}^n(x-u_i)=f$,

and hence $f\in E_H[x]$. Clearly $f(u)=0$, so $p|f$, so $[E:E_H]=\deg p\leq \deg f=n=|H|$, proving the third inequality above. $\Box$

#### The Properties

Property 1. If $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$ and if $K_1\subset K_2$ then $\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_1)$.

Proof of Property 1. Easy. $\Box$

Property 2. $[E:K]=|\operatorname{Gal}(E/K)|$ and $[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.

Proof of Property 2. If $K=E_H$, then $|\operatorname{Gal}(E/K)|=|\operatorname{Gal}(E/E_H)|=[E:E_H]=[E:K]$ as was shown within the proof of $\Phi\circ\Psi=I$. But every $K$ is $E_H$ for some $H$, so $|\operatorname{Gal}(E/K)|=[E:K]$ for every $K$ between $E$ and $F$. The second equality follows from the first and from the multiplicativity of the degree/order/index in towers of extensions and in towers of groups:

$[K:F] = \frac{[E:F]}{[E:K]} = \frac{|\operatorname{Gal}(E/F)|}{|\operatorname{Gal}(E/K)|} = [\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)].\quad\Box$

Property 3. If $K$ in $E/K/F$ is the splitting field of a polynomial in $F[x]$ then $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.

Proof of Property 3. We will define a surjective (onto) group homomorphism $\rho:\operatorname{Gal}(E/F)\to\operatorname{Gal}(K/F)$ whose kernel is $\operatorname{Gal}(E/K)$. This shows that $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ (kernels of homomorphisms are always normal) and then by the first isomorphism theorem for groups, we'll have that $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.

Let $\sigma$ be in $\operatorname{Gal}(E/F)$ and let $u$ be an element of $K$. Let $p$ be the minimal polynomial of $u$ in $F[x]$. Since $K$ is a splitting field, lemma 2 implies that $p$ splits in $K[x]$, and hence all the other roots of $p$ are also in $K$. As $\sigma(u)$ is a root of $p$, it follows that $\sigma(u)\in K$ and hence $\sigma(K)\subset K$. But since $\sigma$ is an isomorphism, $[\sigma(K):F]=[K:F]$ and hence $\sigma(K)=K$. Hence the restriction $\sigma|_K$ of $\sigma$ to $K$ is an automorphism of $K$, so we can define $\rho(\sigma)=\sigma|_K$.

Clearly, $\rho$ is a group homomorphism. The kernel of $\rho$ is those automorphisms of $E$ whose restriction to $K$ is the identity. That is, it is $\operatorname{Gal}(E/K)$. Finally, as $E/F$ is a splitting extension, so is $E/K$. So every automorphism of $K$ extends to an automorphism of $E$ by the uniqueness statement for splitting extensions. But this means that $\rho$ is onto. $\Box$

## Extra Practice Problems with Solutions

Attached were some problems User:Clkkang found on the net with solutions. They may be very useful when studying for problems involving galois theory and other subjects. They were deleted by Dror for fear of copyright violations (it will be appropriate for User:Clkkang to post here links to the original sources of those pages).

The following attachments are notes that User:Clkkang had taken from the text, it "MAY BE" a little easier to read than the text due to the formatting: