Ideal in the ring of symetric polynomials I need to find generators of the ideal $J$ in the ring of symmetric polynomials of 3 variables $(x_1, x_2, x_3)$ such that $\forall j \in J: x_1 = x_2 \Rightarrow 
 j=0$.
It is clear that polynomials $(x_1-x_2)(x_2-x_1)$ and $(x_1 - x_2)^2$ are generators. So are this polynomials enough to generate whole $J$ and if yes, how can I prove it?
Thanks!
 A: Take a symmetric function $f(x_1, x_2, x_3)$. View it as an element of $\mathbf{Q}(x_2,x_3)[x_1]$. For polynomials over a field, we know that $g(t) \in k[t]$ has a root at $a$ if and only if $(t - a) \mid g$.
Thus since $f(x_2,x_2,x_3) = 0$ it must be that $(x_2 - x_1) \mid f$ in $\mathbf{Q}(x_2,x_3)[x_1]$. By Gauss's Lemma, we can say that $(x_2 - x_1) \mid f$ in $\mathbf{Q}[x_1,x_2,x_3]$ (or $\mathbf{Z}[x_1,x_2,x_3]$ if $f$ had integer coefficients).
Therefore write $f = (x_1 - x_2)f_1$. If we now apply the permutation $(1,2)$ we get $f = (x_2 - x_1)\cdot (1,2)f_1$. Hence $(1,2)f_1 = -f_1$. But from this we know that $f_1$ must be zero if $x_1 = x_2$. Therefore $f = (x_1 - x_2)^2f_2$.
Thanks to @darij grinberg for pointing this out: notice that $(x_1 - x_2)^2$ is not symmetric because it isn't invariant under switching $x_1$ and $x_3$ or $x_2$ and $x_3$. Notice that actually the ideal $J$ is the set of polynomials that vanish when $x_1 = x_2$ or $x_1 = x_3$ or $x_2 = x_3$ by symmetry. Thus using the same logic as above, if $f \in J$ then $f = (x_1 - x_2)^2(x_1 - x_3)^2(x_2 - x_3)^2f_3$. So $J$ is generated by $(x_1 - x_2)^2(x_1 - x_3)^2(x_2 - x_3)^2$.
