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Let $J\subseteq k[x_1, \ldots, x_{n-1}]\subseteq k[x_1, \ldots, x_n]$ be an ideal such that $(x_1, \ldots, x_{n-1})$ is a minimal prime of $J$ (thinking in the polynomial ring $k[x_1, \ldots, x_n]$). Is $J$ an ideal $(x_1, \ldots, x_{n-1})$-primary (thinking in $k[x_1, \ldots, x_{n-1}]$)?

J. W. Tanner
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Euler
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  • Surely the same set $J$ is not an ideal in these two rings. I think you should define your setup more explicitly. – Lubin Feb 04 '19 at 04:50
  • Set $J=(a_1, \ldots, a_r)\subseteq k[x_1, \ldots, x_{n-1}]$. Now let $J':=(a_1, \ldots, a_r)$ as an ideal of $k[x_1, \ldots, x_n]$. Suppose $(x_1, \ldots, x_{n-1})\subseteq k[x_1, \ldots, x_n]$ is a minimal prime of $J'$. Is $J$ a $(x_1, \ldots, x_{n-1})$-primary? – Euler Feb 04 '19 at 04:57
  • Well, $(x_1,\cdots,x_{n-1})$ doesn’t look like any kind of minimal prime to me… – Lubin Feb 04 '19 at 05:07
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    What about $J=(X,Y)\cap(X-1,Y)$ in $K[X,Y]$? – user26857 Feb 04 '19 at 09:07

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