Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Emmy Noether's 1915 paper "Der Endlichkeitssatz der Invarianten endlicher Gruppen", I saw the notion of a 'Galois resolvent', which I don't quite understand. Google didn't really help me with that, since it seems to be different from what I have found to this notion. Let me try to translate what happens there:

We have a finite group $H$ of invertible $n\times n$-matrices $A_1,\dots,A_h$, where $A_k=(a_{ij}^{(k)})$, with entries in some field $K$ of characteristic $0$ (I'm actually not 100% sure if this is sufficient, in a later paper she talks about 'übliche Zahlkörper'). Then $H$ acts on the polynomial ring $K[x_1,\dots,x_n]$, where we set $x=(x_1,\dots,x_n)^T$, via

$$(A_k,x)\mapsto A_k\cdot x,$$

and we denote $A_k\cdot x$ also by $x^{(k)}$. Since one of the $A_k$ is the identity, there is a $k$ such that $x^{(k)}=x$.

Now a polynomial invariant of $G$ is some $f\in K[x_1,...,x_n]$ for which $A_k\cdot f=f$ for all $k$. In other words, we have


Now comes the part where I'm stuck, I try to translate it as good as I can:

This formula expresses that $f$ is a a polynomial, symmetric function in the $x^{(k)}$. The theorem about symmetric functions of "Größenreihen" (I don't know how to translate this) says that $f$ can be represented in a polynomial way by the elementary symmetric functions of these "Reihen" (series), that is, by the coefficients $G_{\alpha,\alpha_1,\dots,\alpha_n}(x)$ of the "Galois resolvent":

\begin{align*}\phi(z,u)&=\prod_{k=1}^h(z+u_1x_1^{(k)}+\dots+u_nx_n^{(k)})\\&=z^h+\sum G_{\alpha,\alpha_1,\dots,\alpha_n}(x)z^\alpha u_1^{\alpha_1}\cdots u_n^{\alpha_n}\begin{pmatrix}(\alpha+\alpha_1+\dots+\alpha_n=h)\\\alpha\neq h\end{pmatrix},\end{align*}

where the $G_{\alpha,\alpha_1,\dots,\alpha_n}(x)$ are invariants of degree $\alpha_1+\dots+\alpha_n$ in the $x_i$.

I actually don't have any idea where $u,z$ come from. I thought there was a nice expression for the elementary symmetric polynomials, and that they are coefficients of a much simpler polynomial. The theorem mentioned above isn't referenced further, I guess the main theorem on symmetric polynomials is meant, but I fail to get the connection to this strange formula.

I'd be glad if someone could help me out here, and explain where this 'Galois resolvent' comes from and what's the connection to the 'usual' stuff about symmetric polynomials. Thank you very much in advance!

share|cite|improve this question
+1 interesting... – draks ... Feb 28 '13 at 11:43
The notion of a Galois resolvent is defined in this wikipedia article. – JSchlather Feb 28 '13 at 18:40
Theorem 2 here might be helpful: – marlu Feb 28 '13 at 20:02
@JSchlather Thanks, I'll have to go over that. But do you know why this notion appears here? I mean, from the text Noether seems to say that the $G_{\dots}$ are the elementary symmetric polynomials, which would be much easier to obtain than via this. And still I can't see where the connection to resolvents really comes from here. Thanks for the link! – InvisiblePanda Mar 8 '13 at 9:11
@JSchlather the definition of the Galois resolvent in the wikipedia page you cite is incomprehensible as it stands today - it talks about permutation groups acting on polynomials without being precise about what the coefficient fields are and it mixes up orbits with permutation groups. I don't feel competent to attempt to fix it myself, but it would be nice if it were fixed. – Rob Arthan Jul 19 '13 at 23:27

(adding comment as answer)

The concept of resolvent in Lagrange/Galois's time and in modern time is different. This paper examines the resolvent concept in the time of Lagrange/Galois and its connections with Galois theory.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.