7 edit clearly indicated

Although I'm very interested in the history of Galois Theory, I know almost nothing about it

EDIT. 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 part of the answer that has been rewritten:

We give below a lattice anti-isomorphism.

The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.

The first publications short proof of this formulation by Artin himself were "Foundations the Fundamental Theorem of Galois Theory " (1938) FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and "Galois Theory" (1942).b). These two statements, and the way they are proved here, go back at least to Emil Artin himself doesn't seem (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 ever explicitly claimed this discoveryseen so far are much more complicated.

Assuming all this is trueSo, my 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 (probably naivelink) question is:

Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?

I also hope this of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is not completely unrelated implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the question.

Here splitting field of a polynomial is an exercise I'm sure everybody has done onceverified. It is to give an as short as possible complete statement and Artin's proof shows in fact that, when the roots of the Fundamental 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 (finite) Galois-Artin Theoryusing Artin's argument) was written with Keith Conrad's help.

THEOREM

Theorem. Let $E/F$ be a finite degree an extension of fields. Assume that $E$ is generated over , let $F$ by a_1,\dots,a_n$be distinct elements generators of$a_1,\dots,a_n$E/F$ such that the product $p$ of the $X-a_i$ is in $F[X]$. Then

$\bullet$

• the group $G$ of $F$-automorphisms automorphisms of $E/F$ is finite,

$\bullet$

• there is a bijective correspondence between the subextensions sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have$$S\leftrightarrow S\leftrightarrow H\iff H=Aut_S E\iff S=E^H\Rightarrow [E:S]=|H|,$$ H=\text{Aut}(E/S)\iff S=E^H

PROOF. We will use the following things: the universal property of polynomial rings and that of quotients of rings by ideals, the principality of polynomial rings over fields, the multiplicativity of the degree of (finite degree) field extensions, the relation between orders and indexes for subgroups of finite groups, and the fact that homogeneous systems of linear equations (over a field) with more unknowns than equations have nontrivial solutions.

(a) If $S/F$ is a subextension sub-extension of $E/F$, then $[E:S]=|\mathrm{Aut}_S E|$. [E:S]=|\text{Aut}(E/S)|$. Proof that (a) and (b) imply the Theoremtheorem. Let$S/F$be a subextension sub-extension of$E/F$and put$H:=Aut_S E$. H:=\text{Aut}(E/S)$. 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:=Aut_{E^H}E$H:=\text{Aut}(E/E^H)$. 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 show check that any $F$-morphism F$-embedding$f_i$from \phi$ of $S_i:=S(a_1,\dots,a_i)$ to K$in$E$has exactly$[S_{i+1}:S_i]$[L:K]$ extensions to an $F$-morphism F$-embedding$f_{i+1}$from \Phi$ of $S_{i+1}$ to L$in$E$. Let E$; or, equivalently, that the polynomial $q$ be a generator of p\in\phi(K)[X]$which is the kernel image under$\phi$of the evaluation at$a_{i+1}$from$S_i[X]$to minimal polynomial of$E$. It suffices to check that a_i$ over $f_i\,q$ K$has$\deg q$[L:K]$ distinct roots in $E$. But this follows from the facts that $q$ divides is clear since $p$ and that divides the product of the $f_i\,p=p$. 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|$, and let pick a$b$be in$E^k$. b=(b_1,\dots,b_k)\in E^k.We must show that the$b_i$are linearly dependent over$E^H$, or equivalently that$b^\perp\cap(E^H)^k$is nonzero. Let , where$x$be a vector \bullet^\perp$ denotes the vectors orthogonal to $\bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $(H b)^\perp$ satisfying b^\perp\cap (E^H)^k$is necessarily orthogonal to$x_j=1$hb$ for some any $j$ and h\in H$, so $$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 as many the largest number of i as possible (such exists because among all nonzero vectors in (H b)^\perp (Hb)^\perp. Some coordinate x_j is nonzero). For 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 difference h x-x j-th coordinate of hx-x is in (H b)^\perp, and thus equal to zero (0, so hx-x=0 by the choice of x). This implies that 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. 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. • 6 fixed a broken link 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. Here is an exercise I'm sure everybody has done once. It is to give an as short as possible complete statement and proof of the Fundamental Theorem of (finite) Galois-Artin Theory. THEOREM. Let E/F be a finite degree extension of fields. Assume that E is generated over F by distinct elements a_1,\dots,a_n such that the product p of the X-a_i is in F[X]. Then \bullet the group G of F-automorphisms of E/F is finite, \bullet there is a bijective correspondence between the subextensions S/F of E/F and the subgroups H of G, and we have$$S\leftrightarrow H\iff H=Aut_S E\iff S=E^H\Rightarrow [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 will use the following things: the universal property of polynomial rings and that of quotients of rings by ideals, the principality of polynomial rings over fields, the multiplicativity of the degree of (finite degree) field extensions, the relation between orders and indexes for subgroups of finite groups, and the fact that homogeneous systems of linear equations (over a field) with more unknowns than equations have nontrivial solutions. We claim: (a) If S/F is a subextension of E/F, then [E:S]=|\mathrm{Aut}_S E|. (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 subextension of E/F and put H:=Aut_S E. 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:=Aut_{E^H}E. Then we have trivially H\subset\overline H, and (a) and (b) imply |H|=|\overline H|. Proof of (a). It suffices to show that any F-morphism f_i from S_i:=S(a_1,\dots,a_i) to E has exactly [S_{i+1}:S_i] extensions to an F-morphism f_{i+1} from S_{i+1} to E. Let q be a generator of the kernel of the evaluation at a_{i+1} from S_i[X] to E. It suffices to check that f_i\,q has \deg q distinct roots in E. But this follows from the facts that q divides p and that f_i\,p=p. 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 let b be 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. Let x be a vector of (H b)^\perp satisfying x_j=1 for some j and x_i=0 for as many i as possible (such exists because (H b)^\perp is nonzero). For any h in H the difference h x-x is in (H b)^\perp, and thus equal to zero (by the choice of x). This implies that x is in (E^H)^k. 5 Added exercise (i.e. proof of the FTGT). Here is an exercise I'm sure everybody has done once. It is to give an as short as possible complete statement and proof of the Fundamental Theorem of (finite) Galois-Artin Theory. THEOREM. Let E/F be a finite degree extension of fields. Assume that E is generated over F by distinct elements a_1,\dots,a_n such that the product p of the X-a_i is in F[X]. Then \bullet the group G of F-automorphisms of E/F is finite, \bullet there is a bijective correspondence between the subextensions S/F of E/F and the subgroups H of G, and we have$$S\leftrightarrow H\iff H=Aut_S E\iff S=E^H\Rightarrow [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 will use the following things: the universal property of polynomial rings and that of quotients of rings by ideals, the principality of polynomial rings over fields, the multiplicativity of the degree of (finite degree) field extensions, the relation between orders and indexes for subgroups of finite groups, and the fact that homogeneous systems of linear equations (over a field) with more unknowns than equations have nontrivial solutions. We claim: (a) If$S/F$is a subextension of$E/F$, then$[E:S]=|\mathrm{Aut}_S E|$. (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 subextension of$E/F$and put$H:=Aut_S E$. 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$\overline H:=Aut_{E^H}E$. Then we have trivially$H\subset\overline H$, and (a) and (b) imply$|H|=|\overline H|$. Proof of (a). It suffices to show that any$F$-morphism$f_i$from$S_i:=S(a_1,\dots,a_i)$to$E$has exactly$[S_{i+1}:S_i]$extensions to an$F$-morphism$f_{i+1}$from$S_{i+1}$to$E$. Let$q$be a generator of the kernel of the evaluation at$a_{i+1}$from$S_i[X]$to$E$. It suffices to check that$f_i\,q$has$\deg q$distinct roots in$E$. But this follows from the facts that$q$divides$p$and that$f_i\,p=p$. 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 let$b$be 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. Let$x$be a vector of$(H b)^\perp$satisfying$x_j=1$for some$j$and$x_i=0$for as many$i$as possible (such exists because$(H b)^\perp$is nonzero). For any$h$in$H$the difference$h x-x$is in$(H b)^\perp$, and thus equal to zero (by the choice of$x$). This implies that$x$is in$(E^H)^k\$.

4 corrected spelling