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Assume $F$ is a field and assume $f\in F[x_1,\ldots,x_4]$ is a polynomial that is invariant under the Klein Four group $V_4$. How can I show that this polynomial can then be rewritten as a polynomial $g$ with coefficients of symmetric polynomials $s_1,\dots,s_4$ and in the variables $y_1, y_2, y_3$, where $$ y_1 = x_1 x_2 + x_3x_4, ~~~ y_2 = x_1 x_3 + x_2x_4, ~~~y_3 = x_1 x_4 + x_2x_3, ~~~ $$

so $f(x_1,x_2,x_3,x_4) = g(y_1, y_2, y_3)$.


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Assuming $F$ doesn't have characteristic $2$, change variables to $u_{++} = x_1+x_2+x_3+x_4$, $u_{+-} = x_1+x_2-x_3-x_4$, $u_{-+} = x_1-x_2+x_3-x_4$, $u_{--} = x_1-x_2-x_3+x_4$. Then the action simply multiplies the $u$'s by characters of $V_4$, and it is obvious that the invariant ring is $F[u_{++}, u_{+-}^2, u_{-+}^2, u_{--}^2]$. –  David Speyer Feb 8 '13 at 20:20

3 Answers 3

Unless I'm making a very obvious mistake this is false: $f = x_1 + x_2 + x_3 + x_4$ is invariant under the Klein $4$-group but has degree $1$. Your $y_i$ have degree $2$.

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You are right! @Edward, I believe the correct formulation should require the coefficients of $g$ to be in the field $F(x_1,...,x_4)^G$ of fixed points under the action of the group $G = S_4$. –  Andreas Caranti Feb 8 '13 at 17:10
Or rather the ring of symmetric polynomials. –  Martin Brandenburg Feb 8 '13 at 17:15
I see. Thanks so much for the correction. But I am still not sure how to do it if the coefficients are symmetric polys? –  Edward Feb 8 '13 at 17:17

There is surely an elementary method, but you can use some Galois theory. I am following (by heart, hopefully correctly) the approach of Kaplansky in his Fields and Rings.

Let $E = F(x_1,...,x_4)$, and $K = F(x_1,...,x_4)^G$ be the fixed field under $G = S_4$. Then $\operatorname{Gal}(E/K) \cong S_4$. Note that $G$ permutes the $y_i$ in all 6 possible ways. In particular, $h(x) = (x - y_1)(x - y_2)(x - y_3) \in K[x]$, $L$ is the splitting field of $h(x)$ over $K$, and $\operatorname{Gal}(L/K) \cong S_3$. Now note that the Klein four group fixes each $y_i$, and the Galois correspondence will tell you that $L$ is the fixed field of the Klein four group.

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Interesting. So this answers the question for fields. What about the rings? –  Martin Brandenburg Feb 8 '13 at 17:35
Thanks, @MartinBrandenburg. I'm quite sure it also holds for the rings, but not so sure how to do it. –  Andreas Caranti Feb 8 '13 at 17:40

This is only an idea, too long for a comment. Therefore I make it CW.

It is well-known that $F[x_1,\dotsc,x_n]$ is a free module over its subring of symmetric polynomials $F[x_1,\dotsc,x_n]^{S_n}$. A basis is given by the $n!$ monomials $T_1^{v_1} \cdot \dotsc \cdot T_n^{v_n}$ with $0 \leq v_i < i$.

In our case, $F[x_1,x_2,x_3,x_4]$ is free over $F[x_1,x_2,x_3,x_4]^{S_4}$ of rank $24$ with basis $\{x_2^{v_2} x_3^{v_3} x_4^{v_4} : v_2 \in \{0,1\}, v_3 \in \{0,1,2\}, v_4 \in \{0,1,2,3\}\}$. Write an arbitrary polynomial as

$$p = \sum_{v_2,v_3,v_4} \lambda_{v_2,v_3,v_4} x_2^{v_2} x_3^{v_3} x_4^{v_4}$$

with symmetric polynomials $\lambda_{v_2,v_3,v_4}$. Then $p$ is fix under $V_4 = \langle (1 2)(3 4), (1 3)(2 4) \rangle$ iff we have the following two equations:

$$(1) ~~~~~~~~~~~~ \sum_{v_2,v_3,v_4} \lambda_{v_2,v_3,v_4} x_2^{v_2} x_3^{v_3} x_4^{v_4} = \sum_{v_2,v_3,v_4} \lambda_{v_2,v_3,v_4} x_1^{v_2} x_4^{v_3} x_3^{v_4}$$

$$(2) ~~~~~~~~~~~~ \sum_{v_2,v_3,v_4} \lambda_{v_2,v_3,v_4} x_2^{v_2} x_3^{v_3} x_4^{v_4} = \sum_{v_2,v_3,v_4} \lambda_{v_2,v_3,v_4} x_4^{v_2} x_1^{v_3} x_2^{v_4}$$

In the first equation, write $\lambda_{v_2,v_3,v_4} x_1^{v_2}$ in the basis. After that one can compare coefficients and optains a system of equations for the $\lambda$'s. Similarily for the second equations. One somehow has to solve this ...

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