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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    $\begingroup$ 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 $F[u_{++}, u_{+-}^2, u_{-+}^2, u_{--}^2]$. $\endgroup$ Commented Feb 8, 2013 at 20:20
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    $\begingroup$ @DavidESpeyer: What about the product $u_{+-} u_{-+} u_{--}$? $\endgroup$ Commented Jun 8, 2022 at 15:13
  • $\begingroup$ @darijgrinberg You are right. Does that break the original claim? $\endgroup$ Commented Jun 8, 2022 at 15:17
  • $\begingroup$ Looks like we are okay; that product equals $e_1^3 - 4 e_1 e_2 + 8 e_3$. $\endgroup$ Commented Jun 8, 2022 at 15:21
  • $\begingroup$ What are $e_1,e_2,e_3$? $\endgroup$ Commented Jun 8, 2022 at 15:30

4 Answers 4


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 $2$.

  • $\begingroup$ 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 $G = S_4$. $\endgroup$ Commented Feb 8, 2013 at 17:10
  • $\begingroup$ Or rather the ring of symmetric polynomials. $\endgroup$ Commented Feb 8, 2013 at 17:15
  • $\begingroup$ I see. Thanks so much for the correction. But I am still not sure how to do it if the coefficients are symmetric polys? $\endgroup$
    – Edward
    Commented Feb 8, 2013 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.

  • $\begingroup$ Interesting. So this answers the question for fields. What about the rings? $\endgroup$ Commented Feb 8, 2013 at 17:35
  • $\begingroup$ Thanks, @MartinBrandenburg. I'm quite sure it also holds for the rings, but not so sure how to do it. $\endgroup$ Commented Feb 8, 2013 at 17:40
  • $\begingroup$ @AndreasCaranti one method is the Noether degree bound; I've posted a full solution. $\endgroup$ Commented Oct 8, 2015 at 20:32
  • $\begingroup$ You just use integrality and the fact that there are monic polynomials expressing $x,y,z$ with coefficients in the invariants. $\endgroup$ Commented Oct 9, 2015 at 15:32
  • $\begingroup$ @user142843 - I believe for that to work it would need an argument that the ring $F[s_1,\dots,s_4][y_1,y_2,y_3]$ is integrally closed. (See my longer comment below my own answer for the reason why.) $\endgroup$ Commented Oct 9, 2015 at 16:48

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 ...

  • $\begingroup$ I tried to make this work and got stuck on the fact that I didn't see how to extract a representation in terms of the generators profferred by the OP from the conditions that these equations impose on the $\lambda$'s. However I gave a complete solution below in terms of the Noether degree bound. $\endgroup$ Commented Oct 9, 2015 at 16:50

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.

  • $\begingroup$ When doing Tschinkel's pset couple years found this killer strategy for finding ring of invariants just using that field of rational functions by the field of invariants is Galois, and then using integrality to get the result for rings of invariants. $\endgroup$ Commented Oct 9, 2015 at 15:27
  • $\begingroup$ @user142843 - I don't think that works unless you know for some reason that the ring you're postulating as the invariant ring is integrally closed. For instance you can argue that $k[s_1,\dots,s_n] = k[x_1,\dots,x_n]^{S_n}$ (where $s_i$ are the elem. sym. poly's as in the OP) because $k(s_1,\dots,s_n) = k(x_1,\dots,x_n)^{S_n}$ by Galois, and therefore $k[x_1,\dots,x_n]^{S_n}\subset k(s_1,\dots,s_n)$, and meanwhile $k[x_1,\dots,x_n]^{S_n}\subset k[x_1,\dots,x_n]$ is integral over $k[s_1,\dots,s_n]$, so since $k[s_1,\dots,s_n]$ is already integrally closed we are done. However: $\endgroup$ Commented Oct 9, 2015 at 16:45
  • $\begingroup$ In the present case we'd need to know that $k[s_1,\dots,s_n][y_1,y_2,y_3]$ is integrally closed to apply the same argument. I wouldn't be surprised if it is, but how would you argue that? $\endgroup$ Commented Oct 9, 2015 at 16:46
  • $\begingroup$ (In the case of $k[s_1,\dots,s_n]$ we know it's integrally closed because the algebraic independence of the $s_i$'s implies that it is actually a UFD. But the $y_i$'s are very algebraically dependent on the $s_i$'s, so that doesn't apply in this case.) $\endgroup$ Commented Oct 9, 2015 at 16:52

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