# Passing from Regular sequence to Prime ideal, for power sum symmetric polynomial

Let $S=\mathbb{C}[x_1,x_2,x_3,x_4]$ be a polynomial ring. Let $p_i=x_1^i+\cdots+x_4^i$ be the power sum symmetric polynomial in $\mathbb{C}[x_1,x_2,x_3,x_4]$. Let $I=(p_1,p_2)$ be an Ideal of $\mathbb{C}[x_1,x_2,x_3,x_4]$.

Question: If $(p_1,p_2)$ is a prime ideal, can we also say $(p_m,p_{2m})$ is also a prime ideal in $S$.

Known: We know that $p_1,p_2$ is a regular sequence in $S$. ( Since $p_1,p_2,p_3,p_4$ is a regular sequence in $\mathbb{C}[x_1,x_2,x_3,x_4]$, see Conca, Krattenthaller and Watanabe http://arxiv.org/abs/0801.2662. And subset of a regular sequence is also a regular sequence). By lemma 2.2 in Conca, Krattenthaller and Watanabe, one concludes $p_m,p_{2m}$ is a regular sequence in $\mathbb{C}[x_1,x_2,x_3,x_4]$.

Knowing $(p_1,p_2)$ is a prime ideal, Can one pass to say, $(p_m,p_{2m})$ is also a prime ideal in $S$. To prove $(p_1,p_2)$ is a prime ideal, one can use Serre Criterion, see Theorem 18.15 in Eisenbud, Algebra with a view towards algebraic geometry.

Thanks
Neeraj

• I feel, perhaps, I should add the proof that how the ideal $(p_1,p_2)$ is a prime ideal. This will throw some light as to what the difficulties are, in generalizing. I will add it soon – Neeraj May 27 '12 at 13:32
• But you know $(p_1,p_2)$ is prime right? – J.C. Ottem May 27 '12 at 14:28
• The question seems to be: $(p_1,p_2)$ is a prime ideal (OK). Is there a formal deduction that $(p_m,p_{2m})$ is also a prime ideal(?). – Martin Brandenburg May 27 '12 at 14:29
• For what it's worth, I tried a few small $m < 10$ in a computer (Macaulay2) and for those few small $m$ that I tried, the ideal was still prime. – Karl Schwede May 27 '12 at 23:32
• @ J.C. Ottem: Yes, $(p_1,p_2)$ is a prime ideal. Please see the proof, as below. @ Martin Brandenburg: I feel, one do not have very formal deduction, although, the approach to prove the result is same, only thing is to solve some equations which are arising in similar ways, I was hoping, if possible, one could give a better way to handle those equations, which will prove for all $m$. – Neeraj May 30 '12 at 15:40