# Isomorphism of ideals

I have a question related to this post: http://mathoverflow.net/questions/60412/generic-liftings-of-a-regular-sequence-on-the-initial-ideal

Suppose $I$ and $J$ are ideals in $R=k[x_1,\ldots,x_n]$ with $In(I)\cong In(J)$ where $In(I)$ is the ideal generated by the leading term of all those $f\in I$, with the additional assumption that the minimal number of generators generating $I$ equals the minimal number of generators generating $J$. Then doesn't that imply that $R/I$ is isomorphic to $R/J$?

Can someone give me a counter-example to this or refer me to a reference?

Addendum: suppose $k$ is algebraically closed.

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What does it mean for two ideals to be isomorphic? What do you mean by "the number of generators"? The minimal number? –  Qiaochu Yuan May 22 '12 at 21:29
Thanks Qiaochu. I meant $R/I$ is isomorphic to $R/J$, and yes, I meant the minimal number of generators. I'll make the appropriate edits to the post. –  math-visitor May 22 '12 at 21:32
Also, what do you mean by leading term? Are you fixing a monomial order? –  Qiaochu Yuan May 22 '12 at 21:36
Yes, we fix a monomial order. Any monomial order will do-- we just need to fix it throughout the argument. –  math-visitor May 22 '12 at 21:38
This is quite false even when $n = 2$ and even for ideals which are generated by a single element. Take $I = (x^2 - y), J = (x^2 - y^2)$ where $x > y$ in the monomial order; then $k[x, y]/I \cong k[x]$ but $k[x, y]/J$ has a zero divisor, namely $x + y$.
Edit: Indeed it is false even when $n = 1$. Take $I = (x^2), J = (x^2 - x)$.
Ah, thanks Qiaochu! ${}{}{}$ –  math-visitor May 22 '12 at 21:45