Galois Theory HT25, Fields and field extensions

An injection $\iota : F \hookrightarrow K$.

• $\varphi$ is a ring map (so $F$-linear)
• $\varphi$ respects the injections $\iota _ 1 : F \hookrightarrow K$ and $\iota _ 2 : F \hookrightarrow K’$ (i.e. makes the corresponding diagram commute), so $\varphi \circ \iota _ 2 = \iota _ 1$.

If the field extensions are true subsets, or you are casually identifying $F \subseteq K$ with $\iota(F)$, then a map of $F$-extensions $K \to K’$ is equivalent to an $F$-linear ring homomorphism $K \to K’$.

A map of $F$-extensions $\varphi : K \to K’$ is an $F$-linear ring homomorphism.

There exists a chain of subfields

\[F = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_n\]

such that:

• $\alpha \in F _ n$
• For each $1 \le i \le n$, there exists $\alpha _ i \in F _ i$ and a positive integer $d _ i$ such that $F _ i = F _ {i-1}(\alpha _ i)$ and $\alpha _ i^{d _ i} \in F _ {i-1}$.

Intuitively, it means that $\alpha$ can be written as an expression containing elements of $F$ and using the operations $+$, $-$, $\times$, $\div$ and $\sqrt[k]{\cdot}$.

\[E = F(\alpha)\]

for some $\alpha \in E$.

• $F[\alpha]$ is the subring of $K$ generated by $F$ and $\alpha$
• $F(\alpha)$ is the smallest subfield of $K$ that contains both $F$ and $\alpha$

If $\alpha$ is algebraic over $F$, then $F(\alpha) = F[\alpha]$.

\[F[\alpha] \cong F[t]/\langle m_{F, \alpha} \rangle\]

(why? the first isomorphism theorem for rings applied to the evaluation map $\text{ev} _ \alpha$).

• There exist $\alpha _ 0, \alpha _ 1, \ldots, \alpha _ d \in F$, where $\alpha _ d \ne 0$ and $\alpha _ d y^d + \cdots + \alpha _ 1 y + \alpha _ 0 = 0$
• $\mathbb Q$ has a finite degree extension which contains $y$

If $y$ is not algebraic, then it is transcendental.

@define~

Let $n = [K : F]$, the degree of the field extension. Then $\{1, \alpha, \cdots, \alpha^n\}$ is linearly dependent over $F$. So then there exists $\lambda _ n, \ldots, \lambda _ 0 \in F$ such that

\[\lambda_n \alpha^n + \cdots \lambda_1 \alpha + \lambda_0 = 0\]

So $\alpha$ is algebraic over $F$.

In other words, $K = F(\alpha)$ where $\alpha$ is algebraic. Clearly this extension is simple, and it is finite since $K \cong \frac{F[x]}{\langle f \rangle}$ where $f$ is the minimal polynomial of $\alpha$.

Let $d := \deg m _ {F, \alpha}$. Then $[F[\alpha] : F] = d$.

By the division algorithm, for every $f \in F[t]$, there are unique $q, r \in F[t]$ with $\deg r < d$ such that $f = q m _ {F, \alpha} + r$. Since $\{1, t, \ldots, t^{d-1}\}$ is a basis of $F[t]/\langle m _ {F, \alpha} \rangle$, since $F[t]/\langle m _ {F, \alpha} \rangle \cong F[\alpha]$, it follows that $\{1, \alpha, \ldots, \alpha^{d-1}\}$ is a basis for $F[\alpha]$ over $F$ and so it has degree $d$.

(Hence if $K = F(\alpha)$, then $[K : F] = d$).

\[[F(\alpha_1, \ldots, \alpha_n) : F] < \infty\]

Induct on $n$.

Base case: $n = 1$. This is equivalent to showing that $[F(\alpha _ 1) : F] < \infty$, which follows from the result that says for all $\alpha \in K$ that are algebraic over $F$, $[F(\alpha) : F] = \deg m _ \alpha$.

Inductive step: Suppose that $n > 1$. Then:

\[\begin{aligned} \text{}[F(\alpha_1, \ldots, \alpha_n) : F] &= [F(\alpha_1, \ldots, \alpha_{n-1})(\alpha_n) : F] \\\\ &=[F(\alpha_1, \ldots, \alpha_{n-1})(\alpha_n) : F(\alpha_1, \ldots, \alpha_{n-1})] \times [F(\alpha_1, \ldots, \alpha_{n-1}) :F ] \\\\ &< \infty \end{aligned}\]

where:

• The second equality is justified by the tower law, and
• The final inequality is justified by the inductive hypothesis: on the left for $n = 1$, and on the right for $n-1$.

The $F$-linear ring homomorphisms $\varphi : F(\alpha) \to K$ are in one-to-one correspondence with the roots of $m _ {F, \alpha}$ in $K$.

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Note that $F(\alpha) \cong \frac{F[x]}{\langle m _ {F, \alpha} \rangle}$, and giving a $F$-linear ring homomorphism is the same as choosing where $\alpha$ goes. But the image of $\alpha$ must satisfy the same minimal polynomial, so the maps are in one-to-one correspondence with the roots of $m _ {F, \alpha}$ in $K$.

@important~

This is the same as the number of roots of $x^2 + 1$ which are in $\mathbb C$. Since there are two roots, then there are two maps.

By the tower law

\[[\mathbb Q(\sqrt 2, \sqrt 3) : \mathbb Q] = [\mathbb Q(\sqrt 2, \sqrt 3) : \mathbb Q(\sqrt 2)] \cdot [\mathbb Q(\sqrt 2) : \mathbb Q]\]

Clearly $[\mathbb Q(\sqrt 2) : \mathbb Q] = 2$. To argue that $[\mathbb Q(\sqrt 2, \sqrt 3) : \mathbb Q(\sqrt 2)] = 2$, we need to show that $3$ has no square root in $\mathbb Q(\sqrt 2)$.

Suppose it did so that $(a + b\sqrt 2)^2 = a^2 + 2ab \sqrt 2 + 2b^2 = 3$. If $a, b \in \mathbb Q \setminus 0$, this implies that $\sqrt 2$ is rational, so either $a = 0$ or $b = 0$. If $a = 0$, then $b \ne 0$ and $2 \mid 3$, which is not true. If $b = 0$, then $a \ne 0$ and this implies that $\sqrt 3$ is rational, which is not true.

@example~ @exam~

Squaring it, you get $a + b + \sqrt{ab}$, so $(\alpha^2 - (a + b))^2 - ab$ will send $\alpha$ to zero.

@example~