Who proved the modern form of the fundamental theorem of Galois theory?. Was it in the original Galois' manuscript?
[Math] The fundamental theorem of Galois theory
ho.history-overview
Related Solutions
The development of Galois theory from Lagrange to Artin by B. Melvin Kiernan, is a history of pre-Artin Galois theory.
EDIT. Here is the part of the answer that has been rewritten:
We give below a short proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below).
The derivation of the FTGT from (a) and (b) takes about four lines, but I haven't been able to find these four lines in the literature, and all the proofs of the FTGT I have seen so far are much more complicated. So, if you find either a mistake in these four lines, or a trace of them the literature, please let me know.
The argument is essentially taken from Chapter II (link) of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin's proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin's argument) was written with Keith Conrad's help.
Theorem. Let $E/F$ be an extension of fields, let $a_1,\dots,a_n$ be distinct generators of $E/F$ such that the product of the $X-a_i$ is in $F[X]$. Then
the group $G$ of automorphisms of $E/F$ is finite,
there is a bijective correspondence between the sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have $$ S\leftrightarrow H\iff H=\text{Aut}(E/S)\iff S=E^H $$ $$ \implies[E:S]=|H|, $$ where $E^H$ is the fixed subfield of $H$, where $[E:S]$ is the degree (that is the dimension) of $E$ over $S$, and where $|H|$ is the order of $H$.
PROOF
We claim:
(a) If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|\text{Aut}(E/S)|$.
(b) If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.
Proof that (a) and (b) imply the theorem. Let $S/F$ be a sub-extension of $E/F$ and put $H:=\text{Aut}(E/S)$. Then we have trivially $S\subset E^H$, and (a) and (b) imply $$ [E:S]=[E:E^H]. $$ Conversely let $H$ be a subgroup of $G$ and set $\overline H:=\text{Aut}(E/E^H)$. Then we have trivially $H\subset\overline H$, and (a) and (b) imply $|H|=|\overline H|$.
Proof of (a). Let $1\le i\le n$. Put $K:=S(a_1,\dots,a_{i-1})$ and $L:=K(a_i)$. It suffices to check that any $F$-embedding $\phi$ of $K$ in $E$ has exactly $[L:K]$ extensions to an $F$-embedding $\Phi$ of $L$ in $E$; or, equivalently, that the polynomial $p\in\phi(K)[X]$ which is the image under $\phi$ of the minimal polynomial of $a_i$ over $K$ has $[L:K]$ distinct roots in $E$. But this is clear since $p$ divides the product of the $X-a_j$.
Proof of (b). In view of (a) it is enough to check $|H|\ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a $$ b=(b_1,\dots,b_k)\in E^k. $$ We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^\perp\cap(E^H)^k$ is nonzero, where $\bullet^\perp$ denotes the vectors orthogonal to $\bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $b^\perp\cap (E^H)^k$ is necessarily orthogonal to $hb$ for any $h\in H$, so $$ b^\perp\cap(E^H)^k=(Hb)^\perp\cap(E^H)^k, $$ where $Hb$ is the $H$-orbit of $b$. We will show $(Hb)^\perp\cap(E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^\perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^\perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^\perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j=1$ for some $j$. Since the subspace $(Hb)^\perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $hx\in(Hb)^\perp$, so $hx-x\in(Hb)^\perp$. Since $x_j=1$, the $j$-th coordinate of $hx-x$ is $0$, so $hx-x=0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.
[A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II, available here.
PDF version: http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Selected_Texts/st.pdf
Here is the part of the answer that has not been rewritten:
Although I'm very interested in the history of Galois Theory, I know almost nothing about it. Here are a few things I believe. Thank you for correcting me if I'm wrong. My main source is
http://www-history.mcs.st-and.ac.uk/history/Projects/Brunk/Chapters/Ch3.html
Artin was the first mathematician to formulate Galois Theory in terms of a lattice anti-isomorphism.
The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.
The first publications of this formulation by Artin himself were "Foundations of Galois Theory" (1938) and "Galois Theory" (1942).
Artin himself doesn't seem to have ever explicitly claimed this discovery.
Assuming all this is true, my (probably naive) question is:
Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?
I also hope this is not completely unrelated to the question.
Best Answer
Galois's Proposition I (as translated by Edwards) is:
Let the equation be given whose $m$ roots are $a,b,c,\ldots$. There will always be a group of permuations of the letters $a,b,c,\ldots$ which will have the following property: 1) that each function invariant under the substitutions of this group will be known rationally; 2) conversely, that every function of these roots which can be determined rationally will be invariant under these substitutions.
As Edwards observes, it takes a lot of work to decipher the exact meanings of "substitution" and "invariant" here, but once you've done that, this can be translated into modern language as:
If an element of the splitting field of $K(a,b,c,\ldots)$ is left fixed by all the automorphisms of the Galois group then it is in $K$.
The fundamental theorem of Galois theory (i.e. the Galois correspondence) follows easily, though Edwards doesn't say who first stated it.