Notes - Galois Theory HT25, Finite fields


Flashcards

@State a result about the size of any finite field $F$.


\[|F| = p^n\]

for some prime $p$ and positive integer $n$.

@Prove that if $F$ is a finite field, then $ \vert F \vert = p^n$ for some prime $p$ and some positive integer $n$.


Since $F$ is finite, it must have positive characteristic.

Since $F$ is a field, the characteristic must be a prime $p$.

Hence $F$ contains a copy of $\mathbb F _ p$, generated by the identity.

Considering $F$ as a vector space over $\mathbb F _ p$, note that since $F$ is finite, it must have a finite dimension $n$. Therefore $ \vert F \vert = p^n$.

Suppose $F$ is a field of characteristic $p$. @Define the Frobenius endomorphism $\phi$.


\[\phi : F \to F\]

defined by

\[\phi(x) := x^p\]

for all $x \in F$.

@Prove that the Frobenius endomorphism $\phi$ for a field $F$ of characteristic $p$ is always in the Galois group $\text{Gal}(F/\mathbb F _ p)$ (also proving that it is indeed an endomorphism).


$\phi$ is multiplicative: $\phi(xy) = (xy)^p = x^p y^p$.

$\phi$ is additive: This is the “Freshman’s dream”: by the binomial theorem,

\[(x+y)^p = \sum^p_{i = 0} {p \choose i} x^p y^{p-i}\]

Since $p \mid {p \choose i}$ for all $0 < i < p$, it follows that

\[(x + y)^p = x^p + y^p\]

$\phi$ is injective: Suppose that $\phi(a) = \phi(b)$. Then $\phi(a - b) = 0$. Since $F$ is in particular an integral domain, it follows that $a - b = 0$, and so $a = b$.

$\phi$ is surjective: $F$ is finite, so injectivity implies surjectivity.

$\phi$ is $\mathbb F _ p$-linear: $\phi(x) = x^p = x$ for all $x \in \mathbb F _ p$ by Fermat’s Little Theorem. Hence $\phi \in \text{Gal}(F/\mathbb F _ p)$.

Suppose:

  • $p$ is a prime
  • $n$ is a positive integer

State two results that relate finite fields of size $p^n$ and splitting fields.


  1. Any field of order $p^n$ is a splitting field of $t^{p^n} - t$ over $\mathbb F _ p$.
  2. Any splitting field of $t^{p^n} - t$ over $\mathbb F _ p$ is a field of order $p^n$.

i.e. a field is of order $p^n$ iff it is a splitting field of $t^{p^n} - t$ over $\mathbb F _ p$.

Suppose:

  • $p$ is a prime
  • $n$ is a positive integer
  • $F$ is a field of size $p^n$

@Prove that:

  1. Any field of order $p^n$ is a splitting field of $t^{p^n} - t$ over $\mathbb F _ p$.
  2. Any splitting field of $t^{p^n} - t$ over $\mathbb F _ p$ is a field of order $p^n$.

(1): $F$ must contain a copy of $\mathbb F _ p$.

Then $x^{p^n - 1} = 1$ for all $x \in F^\times$ by Lagrange’s Theorem.

Therefore all elements of $F$ are roots of $t^{p^n} - t$. Therefore $t^{p^n} - t$ splits completely in $F[t]$ and its roots generate $F$. Hence $t^{p^n} - t$ splits completely in $F[t]$, and its roots generate $F$. Hence $F$ is a splitting field of $t^{p^n} - t$ over $\mathbb F _ p$.

(2): Let $F$ be a splitting field of $f = t^{p^n} - t$ containing $\mathbb F _ p$. Suppose that $\alpha$ is a repeated root of $f$ in $F$.

Then $D(f)(\alpha) = 0$. But $D(f) = p^n t^{p^n - 1} - 1 = -1$, a contradiction. Hence $f$ has $\deg(f) = p^n$ distinct roots in $F$, so $ \vert F \vert \ge p^n$.

Let $V$ be the set of roots of $f$ in $F$. Then

\[V = \\{\alpha \in F : \alpha^{p^n} = \alpha\\} = F^{\langle \phi^n \rangle}\]

where $\phi$ is the Frobenius endomorphism. Since $F$ is a splitting field of $f$ by assumption, $V$ generates $F$ as a field, and so $V = F$.

Hence $ \vert F \vert = \vert V \vert = p^n$.

@Prove, by appealing to other results, that:

  1. Up to isomorphism there is a unique field $\mathbb F _ {p^n}$,
  2. $\mathbb F _ {p^n}$ is Galois over $\mathbb F _ p$.

(1): There is a result that says

Any field of order $p^n$ is a splitting field of $t^{p^n} - t$ over $\mathbb F _ p$.

But there is another result that says:

Any two splitting fields of a polynomial are isomorphic.

(2): If $g$ is an irreducible factor of $t^{p^n} - t$, then $g$ has no repeated roots in any splitting field. Therefore $g$ is separable by the result that says

If:

  • $X \subseteq L$

Then the polynomial $f _ X := \prod _ {y \in X} (t - y) \in L[t]$ is always separable.

But by the theorem that says

Any splitting field of $t^{p^n} - t$ over $\mathbb F _ p$ is a field of order $p^n$.

$\mathbf F _ {p^n}$ is a splitting field of the separable polynomial $t^{p _ n} - t$ and is hence Galois over $\mathbb F _ p$.

@Prove that the Frobenius endomorphism

\[\phi : \mathbb F_{p^n} \to \mathbb F_{p^n}\]

has order $n$ in $\text{Gal}(\mathbb F _ {p^n} / \mathbb F _ p)$.


Since $\alpha^{p^n} = \alpha$ for all $\alpha \in \mathbb F _ {p^n}$ by Fermat’s Little Theorem (Lagrange?), it follows that $\phi^n = 1$.

Suppose that $\phi^m = 1$ for some $1 \le m \le n$.

Then $\alpha^{p^m} = \alpha$ for all $\alpha \in \mathbb F _ {p^n}$. But then $t^{p^m} - t$ has $ \vert \mathbb F _ {p^n} \vert $ distinct roots in $\mathbb F _ {p^n}$, which implies that $p^m \ge p^n$. Hence $m \ge n$, and the order of $\phi$ is precisely $n$.

@State a result about the Galois group

\[\text{Gal}(\mathbb F_{p^n} / \mathbb F_p)\]

It is cyclic of order $n$, and is generated by the Frobenius endomorphism $\phi : \mathbb F _ {p^n} \to \mathbb F _ {p^n}$.

@Prove that

\[\text{Gal}(\mathbb F_{p^n} / \mathbb F_p)\]

is a cyclic group of order $n$ and is generated by the Frobenius endomorphism $\phi : \mathbb F _ {p^n} \to \mathbb F _ {p^n}$.


By the result that says

Suppose $K/F$ is a finite field extension and $G = \text{Gal}(K/F)$. Then $ \vert G \vert \le [K : F]$.

it follows that

\[|\text{Gal}(\mathbb F_{p^n}/\mathbb F_p)| \le [\mathbb F_{p^n} : \mathbb F_p] = n\]

but then by the result that says $\phi$ has order $n$, it must actually be the case that

\[|\text{Gal}(\mathbb F_{p^n} / \mathbb F_p)| = n\]

and $\phi$ generates all of $\text{Gal}(\mathbb F _ {p^n} / \mathbb F _ p)$.

@State a result about when $\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$ (i.e. there is an injective field homomorphism from $\mathbb F _ {p^a}$ to $\mathbb F _ {p^b}$).


$\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$ if and only if $a \mid b$.

@Prove that $\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$ if and only if $a \mid b$.


Forward direction: Suppose that $\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$. Let $G = \text{Gal}(\mathbb F _ {p^b}/\mathbb F _ p)$, then $G$ is cyclic by the result that says

The Galois group $\text{Gal}(\mathbb F _ {p^n} / \mathbb F _ p)$ is cyclic of order $n$, and is generated by the Frobenius endomorphism $\phi : \mathbb F _ {p^n} \to \mathbb F _ {p^n}$.

and has order $b$. Let $H = \text{Gal}(\mathbb F _ {p^b} / \mathbb F _ {p^a})$ be the subgroup of $G$ corresponding to $\mathbb F _ {p^a}$ (which exists by the Galois correspondence). Since $G$ is abelian, $H$ is normal and $G/H \cong \text{Gal}(\mathbb F _ {p^a} / \mathbb F _ p)$. But then by the above result, $G/H$ is also a cyclic group of order $a$.

But then $a \mid b$ by Lagrange’s theorem.

Backward direction: Suppose that $a \mid b$. Then $p^a - 1 \mid p^b - 1$ (this can be seen by considering $b = ka$ and then expanding as a sum of powers). Then similarly,

\[t^{p^a - 1} - 1 \mid t^{p^b - 1} - 1\]

Multiplying through by $t$, we see

\[t^{p^a} - t \mid t^{p^b} - t\]

Therefore $t^{p^a} - t$ splits completely in $\mathbb F _ {p^b}[t]$ (since previous results tell us that $t^{p^b} - t$ splits completely in this field). But then it follows that $\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$ by the result that says

Let $f \in F[t]$ be a separable polynomial, and let $K$ be a splitting field for $f$. Suppose that $L$ is another extension of $F$ such that $f$ splits completely in $L[t]$. Then there exists at least one injective homomorphism $K \hookrightarrow L$.

it follows that $\mathbb F _ {p^a}$ embeds into $\mathbb F _ {p^b}$.




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