Let $G$ be a finite group with $n$ elements and let $\mathbb{Q}[x_1,\cdots,x_n]^G = \mathbb{Q}[g_1,\cdots,g_m]$. Then there exist polynomials $s_j \in \mathbb{Q}[y_1,\cdots,y_m]$ for $j=1,\cdots,n$ such that $e_j(x_1,\cdots,x_n) = s_j(g_1(x_1,\cdots,x_n), \cdots, g_m(x_1,\cdots,x_n))$ for $j=1,\cdots,n$ where $e_j$ is the $j$-elementary symmetric polynomial in $x_k$. Consider the polynomial $p(t,y_1,\cdots,y_m) = t^n - s_1(y_1,\cdots,y_m) t^{n-1}+\cdots+(-1)^n s_n(y_1,\cdots,y_m)$ which is a polynomial in $R:=\mathbb{Q}[t,y_1,\cdots,y_m]$. Is it true, that this polynomial is irreducible in $R$?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ How is the action of $G$ defined? $\endgroup$– Francesco PolizziCommented Jan 9, 2017 at 14:06
-
$\begingroup$ Consider the regular representation of G. And let the action of G be defined through the regular representation. $\endgroup$– user6671Commented Jan 9, 2017 at 14:08
-
1$\begingroup$ I am a bit unclear about what this question says. I want to say that once the dust clears one should be able to attack it in a relatively straightforward manner using Galois theory but when I tried to do this I got a bit confused. What does the notation $\mathbb{Q}[g_1,g_2,\ldots,g_m]$ mean? Are the $g_i$ elements of $\mathbb{Q}[x_1,x_2,\ldots,x_n]^G$ with the property that the $\mathbb{Q}$-algebra generated by the $g_i$ is $\mathbb{Q}[x_1,\ldots,x_n]^G$ or do you want more than this? If there is some flexibility for the $g_i$ then I guess there's a chance that the answerDependsOnMoreThan $G$? $\endgroup$– Kevin BuzzardCommented Jan 9, 2017 at 22:46
-
$\begingroup$ I mean it like this: Every invariant polynomial $f \in \mathbb{Q}[x_1,\cdots,x_n]^G$ can be written as a polynomial $f = s(g_1,\cdots,g_m)$ in $g_j$ where $s \in \mathbb{Q}[y_1,\cdots,y_m]$. $\endgroup$– user6671Commented Jan 10, 2017 at 7:16
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
I think it's irreducible in $R$ as you suggest. Here's a sketch which I think works. If the polynomial factored in a non-trivial way, then because of the $t^n$ term the factors must have degree less than $n$ in $t$ (consider the factorization in $\mathbb{Q}(y_1,y_2,\ldots,y_n)[t]$; note also that we can assume that the factors over $R$ are monic polynomials in $t$). Now specialise via $y_i\mapsto g_i$ and we get a non-trivial factorization of the specialised polynomial in $\mathbb{Q}[g_1,\ldots,g_m][t]$ and hence in $\mathbb{Q}[x_1,\ldots,x_n][t]$ and so in $\mathbb{Q}(x_1,\ldots,x_n)[t]$. But we know the complete factorization in this ring, it's just $\prod(t-x_i)$, so our given factorization must specialise into factors of the form $\prod_{i\in I}(t-x_i)$ for some subsets $I$ of $\{1,2,\ldots,n\}$ (with each $I$ not empty or the whole thing), and the constant term of each factor must be in $\mathbb{Q}[g_1,\ldots,g_m]$ and hence $G$-invariant. This is a contradiction.
answered Jan 10, 2017 at 9:08
-
$\begingroup$ First thanks for your answer! I have two questions: 1) Why is the constant term of each factor in $\mathbb{Q}[g_1,\cdots,g_m]$. 2) To what is this a contradiction? $\endgroup$– user6671Commented Jan 10, 2017 at 9:15
-
$\begingroup$ The factorization takes place in $R$ and then you specialise by sending $y_i$ to $g_i$ and then the constant terms are in $\mathbb{Q}[g_1,..,g_m]$ by definition. This is a contradiction because we know from the factorization in the bigger ring $\mathbb{Q}[x_1,...,x_n]$ that the constant terms must be of the form $\prod_{j\in I}x_j$ and these are only in $\mathbb{Q}[g_1,\ldots,g_m]$ for $I$ empty or the whole thing. $\endgroup$ Commented Jan 10, 2017 at 10:21
-
$\begingroup$ Ok, thanks, I will have to work this out in detail. But it makes sense! Thanks again! $\endgroup$– user6671Commented Jan 10, 2017 at 10:24
-
$\begingroup$ This seems to work not just for the regular representation but for $G$ any transitive permutation group of degree $n$ (if we drop the assumption that $n$ is $G$'s order). $\endgroup$ Commented Jun 28, 2017 at 0:05