07-401/Class Notes for April 11

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In Preparation

The information below is preliminary and cannot be trusted! (v)

The Fundamental Theorem of Galois Theory

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 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 $E_{H}$:

$\Psi :\quad H\mapsto E_{H}$.

Furthermore, 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 $\operatorname {Gal} (E/K_{1})>\operatorname {Gal} (E/K_{1})$.
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 a splitting field 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 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

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 of course implies 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 $u_{i}:=\sigma _{i}u$ (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. $\Box$

Proof of Property 1.

Proof of Property 2.

Proof of Property 3.