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Several months ago I asked a question about Cancellation Law of Ideal Multiplication. I believe that the question was too vague and the setting was too general. Now I am interested in learning the following specific case:

Let $R$ be the commutative local ring $R=\mathbb{Z}_p[[T_1,T_2]]$, and $I$ be the prime ideal $I=\langle T_1, T_2\rangle$. Suppose $X$ and $Y$ are two ideals of $R$ such that $X\subseteq Y$. Suppose we know that $I\cdot X=I\cdot Y$ (the usual ideal multiplication), can we possibly conclude that $X=Y$?

Any comment is appreciated. Thank you very much!

BenjaminY
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1 Answers1

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Let $X=\langle T_1^2+T_2^2, T_1T_2\rangle$ and $Y=I^2$.

Clearly $X\subseteq Y$ and $IX=IY$.

To see that $X\neq Y$ just note that $R/X$ is infinite dimensional over $\mathbb{Z}_p$ (indeed its elements are uniquely represented by $p+qT_2$, where $p,q$ are power series in $T_1$), whilst $R/Y$ is 3-dimensional over $\mathbb{Z}_p$.

tkf
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