Jean-Benoît Bost told me a beautiful proof of the main ingredient in the proof of Galois correspondence, which had been published by Lagrange in his 1772 “Réflexions sur la résolution des résolutions algébriques”, almost 60 years before Galois. (See Section 4 of that paper, I think; it is often difficult to recognize our modern mathematics in the language of these old masters.)

In modernized notations, Lagrange considers the following situation. He is given a polynomial equation $ T^n + a_{n-1} T^{n-1}+\cdots + a_0 = 0$, with roots $x_1,\dots,x_n$, and two “rational functions” of its roots $f(x_1,\dots,x_n)$ and $\phi(x_1,\dots,x_n)$. (This means that $f$ and $\phi$ are the evaluation at the $n$-tuple $(x_1,\dots,x_n)$ of two rational functions in $n$ variables.) Lagrange says that $f$ and $\phi$ are similar (“semblables”) if every permutation of the roots which leaves $f(x_1,\dots,x_n)$ unchanged leaves $\phi(x_1,\dots,x_n)$ unchanged as well (and conversely). He then proves that $\phi(x_1,\dots,x_n)$ is a rational function of $a_0,\dots,a_{n-1}$ and $f(x_1,\dots,x_n)$.

Let us restate this in a more modern language. Let $K\to L$ be a finite Galois extension of fields, in the sense that $K= L^{G}$, where $G=\mathop{\rm Aut}_K(L)$. Let $a, b\in L$ and let us assume that every element $g\in G$ which fixes $a$ fixes $b$ as well; then Lagrange proves that $b\in K(a)$.

Translated in our language, his proof could be as follows. In formula, the assumption is that $g\cdot a=a$ implies $g\cdot b=b$; consequently, there exists a unique *function* $\phi\colon G\cdot a\to G\cdot b$ which is $G$-equivariant and maps $a$ to $b$. Let $d=\mathop{\rm Card}(G\cdot a)$ and let us consider Lagrange's interpolation polynomial —the unique polynomial $P\in L[T]$ of degree $d$ such that $P(x)=\phi(x)$ for every $x\in G\cdot a$. If $h\in G$, the polynomial $P^h$ obtained by applying $h$ to the coefficients of $P$ has degree $d$ and coincides with $\phi$; consequently, $P^h=P$. By the initial assumption, $P$ belongs to $K[T]$ and $b=P(a)$, hence $b\in K(a)$, as claimed.

Combined with the primitive element theorem, this allows to give another short, and fairly elementary, presentation of the Galois correspondence.

## Monday, March 23, 2015

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