We understand the Galois group of a polynomial as automorphisms of the polynomial's splitting field. How did Galois, who did not have a notion of "field," much less "automorphism" characterize Galois groups? I understand that it was as a subgroup of the permutation of the roots, but what determined whether a given permutation was in the "Galois group" or not? (I posted this in the Math group on FB, and figured out an answer, but realized that this is the logical place to post such questions. I'll put my answer below, but I'll wait before accepting it in case someone can give a better one.)

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    $\begingroup$ Actually, a better place for you to post this question is here. $\endgroup$ Dec 23 '18 at 22:00
  • $\begingroup$ Thanks, next time I will know. $\endgroup$
    – Nat Kuhn
    Dec 23 '18 at 22:25
  • $\begingroup$ According to the Wikipedia article on Galois he published a paper on finite fields "..., in which the concept of a finite field was first articulated." $\endgroup$
    – Somos
    Dec 24 '18 at 1:15

Michael Livshitz pointed me to this article by Harold M Edwards, according to which the answer is more-or-less this:

Let $A=\{a_1,...,a_n\}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $\pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if):

(g) for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 \iff g(\pi(a_1),...,\pi(a_n))=0$.

It is quite easy to show that this is equivalent to the modern definition. An automorphism of the splitting field of $f$ which leaves $k$ fixed gives a permutation of the roots, and (g) follows directly from the fact that it fixes $k$.

To see the converse, note that the set of all polynomials $g$ in $R$ with $g(a_1,....,a_n)=0$ forms an ideal $I\subset R$; it is just the kernel of the (surjective) map of $k[x_1,...,x_n]$ to $K=k(a_1,...,a_n)$ which takes $g$ to $g(a_1,...,a_n)$. As a result, $R/I$ is isomorphic to $K$. (In fact, $I$ is a maximal ideal but that plays no role in our proof.) If $\pi$ is a permutation satisfying (g), the map from $R$ to $K$ which takes $g$ to $g(\pi(a_1),...,\pi(a_n))$ is also a surjective ring homomorphism, and (g) tells us that it has the same kernel, $I$. So it induces an isomorphism of $K=R/I$ to itself, obviously fixing $k$, and permuting the roots as specified by $\pi$.


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