Ring of invariants of Klein Four group 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 $f(x_1,x_2,x_3,x_4) = g(y_1, y_2, y_3)$.
Thanks.
 A: 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 $2$.
A: 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.
A: 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 ...
A: If the characteristic of $F$ is not 2, here is a solution:
The Noether degree bound tells us that the invariant ring of $V_4$ is generated as an algebra by the homogeneous invariants of degree $\leq 4 = |V_4|$. It is not hard to list all such invariants just by summing over orbits of $V_4$ in the set of monomials:
Degree 1: Since $V_4$ acts transitively, $x_1+\dots+x_4$, i.e. the elementary symmetric polynomial $s_1$, is the only one!
Degree 2: We have $f_2 = x_1^2 + \dots + x_4^2$ (I will use $f_k$ to denote the $k$th power sum), and the three invariants $y_1, y_2, y_3$ that you mentioned.
Degree 3: $f_3$; $s_3$; and three invariants that look like $x_1^2x_2 + x_1x_2^2 + x_3^2x_4 + x_3x_4^2$ (conjugate up to some permutation of $1,2,3,4$).
Degree 4: $f_4$; $s_4$; three invariants that look like $x_1^2x_3x_4 + x_2^2x_3x_4 + x_1x_2x_3^2 + x_1x_2x_4^2$; three more that look like $x_1^3x_2 + x_1x_2^3 + x_3^3x_4+x_3x_4^3$; and finally three that look like $x_1^2x_2^2 + x_3^2x_4^2$.
(In each degree $d$, one gets a conjugacy class of invariants for each partition of $d$, corresponding to the exponents in the monomials. For example, the invariant $x_1^2+x_2^2+x_3^2x_4^2$ and its conjugates correspond to the partition $4=2+2$.)
At any rate, this is a complete list of generators by the Noether bound, so the problem is reduced to the finite calculation of showing that $s_1,\dots,s_4$ and $y_1,y_2,y_3$ generate everything on this list. Indeed, $f_2,f_3,f_4$ are taken care of by the fundamental theorem on symmetric polynomials, and the rest are a fun exercise.
