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If $A \subset B$ is an integral extension, then $\frac{A}{I} $ is integral over $\frac{B}{J}$ where $I=J \cap A$.

I was able to prove this but I am wondering why can't be generalise it for any ideal of $A$?

I was looking for some counterexamples in this case.

Can I try with $I=\mathbb{Q} \subset \mathbb{Q}=A$ and $J={0} \subset \mathbb{Q}(\sqrt2)$

In this case we have $\frac{A}{I}=0$ and $\frac{B}{J}=B$ and definitely $B$ isn't integral over $0$ because there doesn't even exists any non-constant polynomial.

I would like to see some more interesting examples

permutation_matrix
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1 Answers1

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In your example, $A/I$ is not (isomorphic to) a subring of $B/J$. In fact, there isn't even a ring homomorphism $A/I \to B/J$. The hypothesis $I \subset J$ is sufficient to give us a well-defined ring map $A/I \to B/J$ as $a + I \mapsto a + J$.
If we further assume $I = J \cap A$, then the above map is an inclusion.

More generally, one can talk about a ring homomorphism $f : R \to S$ being integral. (This simply means that $S$ is integral over the subring $f(R)$.)

Under this definition, it would make sense to ask if $B/J$ is integral over $A/I$ under just the assumption $I \subset J$. (I am still assuming that $A \subset B$.)
You can check that the usual proof still works.

Aryaman Maithani
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  • when you take $I \subset A \cap J$, then it is inclusion. So probably here we are checking that the induced map need not be injective hence it might not be injection. But if we take $I=A \cap J$ then it will be an injective homomorphism. – permutation_matrix Nov 28 '22 at 10:29