I know that tensor product does not preserves inclusion in general, but is there any concrete example of ideals in $\mathbb{Z}[x_1,...,x_n]$, say $i:I \hookrightarrow J$ such that $i \otimes \mathbb{Z}/p\mathbb{Z}:I \otimes \mathbb{Z}/p\mathbb{Z} \rightarrow J \otimes \mathbb{Z}/p\mathbb{Z}$ is not an inclusion?
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1$\begingroup$ The condition is equivalent to $pI=I \cap pJ$. So for instance $J=\mathbb{Z}[x]$, $I=(p,x)$ is an offender. $\endgroup$– AphelliJun 10, 2022 at 8:42
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$\begingroup$ @Aphelli Isn't $pI = (p^2, px)$ and $I \cap pJ = (p)$ ? $\endgroup$– MathChopperJun 10, 2022 at 9:51
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$\begingroup$ Exactly! Which means that the class of $p$ is nonzero at the beginning but zero at the end. $\endgroup$– AphelliJun 10, 2022 at 10:13
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$\begingroup$ @Aphelli what's an explicit element of the kernel?? I believe $I\otimes\mathbb Z/p\to J\otimes\mathbb Z/p$ is injective. $\endgroup$– Kenta SJun 10, 2022 at 10:18
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$\begingroup$ Nevermind, $p\in I$ is in the kernel. $\endgroup$– Kenta SJun 10, 2022 at 10:22
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