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Let $\mathbb A\subset\mathbb{k}[x_1, .., x_n]$ be the ring of symmetric polynomials, and $\mathcal{J}\subset\mathbb A$ is the ideal of polynomials that vanishes on $x_1=x_2=\cdots=x_k$. How can I find generators of $\mathcal{J}$?

For example, let $n=3$ and $k=2$. If $p(x_1, x_2, x_3)$ vanishes at $x_1=x_2$ then it should be divided by $x_1-x_2$. I also thought about the geometric interpretation (in this example: the zeros of two lines in $R^3$). But I don't understand how to take into account that these polynomials are symmetric.

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I assume that $J$ is meant to be $I(A)$, where $A=\{(x_1,\dotsc,x_n) \in \mathbf{k}^n : x_1=\dotsc=x_k\}$ for some $k \leq n$. But then $A=V(P)$, where $P=\langle X_1-X_2,X_2-X_3,\dotsc,X_{k-1}-X_k\rangle \subseteq \mathbf{k}[X_1,\dotsc,X_n]$. Notice that $P$ is a prime ideal because the quotient ring is isomorphic to $\mathbf{k}[X_k,X_{k+1},\dotsc,X_n]$. So if $\mathbf{k}$ is algebraically closed, we have $J=I(V(P))=P$. A direct calculation shows that this also holds if $\mathbf{k}$ is any field.

Here I use the polynomial ring since you have used the notation for a polynomial ring. The ring of symmetric polynomials is a polynomial ring, the variables are the elementary symmetric polynomals.

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  • $\begingroup$ I didn't quite understand this method. For example, let $n=k=3$. Then $<e_3-e_2, e_2-e_1>$ are generators? But then zeros will not only be the line $x_1=x_2=x_3$. $\endgroup$ – iou Apr 25 '17 at 13:10

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