I'm reading the book Algebra by Lang. In the page 338, it is written that:
Let $A$ be a subring of $B$. Let $p$ be a prime ideal of $A$ and let $P$ be a prime ideal of $B$. We say that $P$ lies above $p$ if $P\cap A=p$.
- I cannot understand this sentence: If that is the case, then the injection $A\rightarrow B$ induces an injection of the factor ring:
$$A/p \rightarrow B/P$$
- Also, if $B$ is integral over $A$, then $B/P$ is integral over $A/p$.
For question 1, I us the map $$\tilde{\phi}:A/p \rightarrow B/P$$ by $\tilde{\phi}(a+p)=\phi(a)+P$ where $\phi$ is the injection $A \rightarrow B$. Then I want to show that if $\phi(a_1)-\phi(a_2)\in P$ then $a_1-a_2\in p$. However $\phi$ is not homomorphism, even for the homomorphism case I couldn't show the injectivity.
For question 2, it refers to prop 1.5 of the book which says that if B be integral over A and $\sigma$ be homomorphism then $\sigma(B)$ is integral over $\sigma(A)$. It's not obvious for me how I can use this lemma.
Consider the composition of maps $A\hookrightarrow B \to B/P$. This is a homomorphism, since it is a composition of homomorphisms. What is the kernel of this ring homomorphism?
Well, it consists precisely of $A\cap P$. But this is $\mathfrak{p}$. Therefore, by the Fundamental Theorem of Homomorphisms, this composition induces an injection $A/\mathfrak{p}\to B/P$.
What is the definition of this embedding? Precisely the one you give! So why are you asserting that $\phi(a+\mathfrak{p}) = \phi(a)+P$ is not a homomorphism?
Note that all the work is done by the First Isomorphism Theorem.
As for the second question, if $B$ is integral over $A$, then $\pi(B)$ is integral over $\pi(A)$, where $\pi\colon B\to B/P$ is the canonical projection. But $\pi(B)=B/P$, and $\pi(A) = AP/P\cong A/(A\cap P) = A/\mathfrak{p}$. That is, $B/P$ is integral over $A/\mathfrak{p}$ (identified as a subring of $B/P$ via the embedding given above).