I am reading Keith Conrad's fantastic notes on the existence of algebraic closures:
https://kconrad.math.uconn.edu/blurbs/galoistheory/algclosureshorter.pdf
My question only concerns the material contained within the first page and the first two sentences of the top of page 2 in the above document.
At the top of page 2, Professor Conrad says (italics added are mine): "Since $I$ is a proper ideal, Zorn's lemma guarantees that $I$ is contained in some maximal ideal $\mathfrak{m}$ in $A$ (Probably $I$ itself is not a maximal ideal, but I don’t have a proof of that.)".
Question. Does anyone know how to prove that the ideal $I$ is not a maximal ideal in the giant polynomial ring $A[X] = K[\{t_{i,f}\}][X]$?
Thanks!
Let $a,b,c \in K$ be distinct. Let $f(x)=(x-a)(x-b)(x-c)$. Then there is an automorphism $\sigma$ of $A/I$ such that: if $g \neq f$, $\sigma(t_{g,i})=t_{g,i}$, and $\sigma$ permutes the $t_{f,i}$ cyclically.
$\sigma$ is not trivial because the $t_{f,i}$ are distinct mod $I$ (indeed, modulo a maximal ideal, they are mapped to $a,b,c$).
Then $\sigma \in Aut_K(A/I)$ has order three. But it’s well-known that the only elements of finite order in an absolute Galois group are the involutions and the unit. None of which has order $3$. So $A/I$ cannot be the algebraic closure of $K$.
(To deal with the case $K=\mathbb{F}_2$, note that we could instead reason with an irreducible monic separable polynomial $f(x)$ of degree $3$ and nothing would change).
Edit: I’m adding a sketch of proof of the claim above (I couldn’t find a reference).
Claim 1: Let $p$ be a prime and $L$ be a field of characteristic not $p$. Assume that $L/K$ is a cyclic Galois extension of degree $p$ and that $L$ contains the $p$-th roots of all its elements. Then $K$ does not contain a primitive $p^2$-th root of unity.
Proof: By assumption, $L$ contains all the $p^{n}$-th roots of unity for all $n$, and thus by Kummer $L$ has no Galois extension whose group is a $p$-group.
Assume that $K$ contains one primitive $p^2$-th root of unity. Then $\mu_{p^2} \subset K$. Now let $q \in \{p,p^2\}$ and consider the following exact sequence over $Gal(L/K)$: $0 \rightarrow \mu_q \rightarrow L^{\times} \rightarrow L^{\times} \rightarrow 0$. Applying cohomology and Hilberg 90 we get $\mathbb{Z}/p\mathbb{Z} \cong K^{\times}/K^{\times q}$.
In particular, $K^{\times p}=K^{\times p^2}$. Thus $K \subset \mu_p K^{\times p} \subset K^{\times p}$ so $K=K^{\times p}$, a contradiction.
Corollary: the same assumptions imply that $p=2$, $L$ and $K$ have characteristic zero, the inclusion $\{\pm 1\} \rightarrow K^{\times}/K^{\times 2}$ is an isomorphism, and $K^{\times 2}$ is stable under sum.
Proof: if $p$ is odd, consider the $p$-adic cyclotomic character $Gal(L/K) \rightarrow \mathbb{Z}_p^{\times}$. Its image is nontrivial (since $\mu_{p^2} \not\subset K$) and contained in the $p$-torsion of $\mathbb{Z}_p^{\times}$: this forces $p=2$ and the character’s image to be $\{\pm 1\}$. We already know (from the proof above) that $-1$ can’t be a square in $K$ and $K^{\times}/K^{\times 2}$ is cyclic.
Apply Tate cohomology of $Gal(L/K)$ to the exact sequence $0 \rightarrow \mu_2 \rightarrow L^{\times} \rightarrow L^{\times} \rightarrow 0$: it implies that $-1$ is the norm of an element of $L$ iff it is the norm of an element of $\mu_2$, which isn’t the case.
So $-1$ isn’t the sum of two squares of $K$, which elementarily implies the rest of the claims.
Corollary: let $L/K$ be a finite extension of fields of characteristic $q$. (We allow $q=0$ even though the notation suggests that $q$ is a prime). Assume that $L$ has no finite abelian (hence, solvable) Galois extension. Then the only primes dividing $|Gal(L/K)|$ are $2$ and $q$ (if $q >0$). Moreover, if $q \neq 2$, the $2$-Sylow of $Gal(L/K)$ is trivial or $\mathbb{Z}/2\mathbb{Z}$ (and in the latter case, $q=0$).
Proof: Using the Galois correspondence, we have two cases to deal with (up to enlarging $K$): $Gal(L/K)$ is cyclic of odd prime order $p\neq q$, and $Gal(L/K)$ is a $2$-group (with $q \neq 2$). The results above rule out the first case and impose, in the second one, the further condition that $q=0$ and the kernel of $2$-adic cyclotomic character must not contain any element of order $2$. So the kernel of said character is trivial, QED.
Claim 2: let $p$ be a prime and $L/K$ be a cyclic Galois extension of degree $p$, with $K$ of characteristic $p$. Then $L$ has cyclic Galois extensions of degree $p$.
Proof: assume that this is false. The idea of the proof is similar to Claim 1, but we use Artin-Schreier theory instead. Let $f(x)=x^p-x$.
We then know that $f(L)=L$. Let’s apply cohomology of $Gal(L/K)$ to $0 \rightarrow \mathbb{F}_p \rightarrow L \rightarrow L \rightarrow 0$ (where the nonzero maps are inclusion and $f$). It follows that (by the normal basis theorem) $\mathbb{F}_p \cong H^1(Gal(L/K),\mathbb{F}_p) \cong K/f(K)$. In other words, $K/f(K)$ has dimension $1$ over $\mathbb{F}_p$.
Write $L=K(a)$. Then it’s easy to see that $f(a^lK)\subset a^lf(K) +\sum_{m <l}{a^m K}$. Thus $f(L) \subset a^{p-1}f(K)+\sum_{m < p-1}{a^mK} \neq L$, netting us a contradiction.
Corollary: let $L/K$ be finite Galois where $L$ has no finite cyclic (hence, no finite solvable) Galois extension. Then, if $L\neq K$, $Gal(L/K)$ has order two, $K$ has characteristic zero, $K^{\times}=\{\pm 1\}K^{\times 2}$, $K^{\times 2}$ is stable under sum and $-1 \notin K^{\times 2}$.