Conceptual reason for some formal properties of integral morphisms of rings

Question

In chapter 5 of Atiyah-Macdonald there are exercises asserting that integral morphisms of rings / integral algebra are stable under tensoring (ex 3) and closed under products (ex 6).

The only definition I know of an integral ring morphism is simply that the inclusion of its image be an integral ring extension, which is in turn defined by monic polynomials.

The formal properties described in Atiyah-Macdonlad seem to hint at a nice functorial description of integral morphisms/extensions.

Is there such a description? If not, what are the conceptual reasons for this behavior?

Answer 1

The stability of integral morphisms under pushout (tensor product) follows from the following two facts.

Proposition. The following is an orthogonal factorization system on the category of algebra over a commutative ring. $$\left(\substack{\text{integral}\\ \text{morphisms} } ,\substack{\text{integrally}\\ \text{closed}\\ \text{monomorphisms} } \right)$$

Proposition. The left class of an orthogonal factorization system is stable under pushouts.

I do not know "why" integral morphisms are closed under products as well.

Question. Are integrally closed monos stable under pushout?

Answer 2

If $A\to B$ is integral, and $C$ is an $A$-algebra, we get a morphism $C\to C\otimes_A B$, which is also integral. To see this, notice that to show a ring map is integral, it suffices to show that a set of algebra generators is integral. Clearly, elements of the for $1\otimes b$, $b\in B$ generates $C\otimes_A B$ as a $C$-algebra. By assumption, there is a monic polynomial $P(x)\in A[x]$ such that $p(b)=0$. Then, $P(x)\in C[x]$ and $P(1\otimes b)=0$, proving what you need. The other part is similar.