0

I have one basic question in the accepted answer
given here In an extension of finitely generated $k$-algebras the contraction of a maximal ideal is also maximal and I don't have enough reputation to comment so I'm posting my question here:

Given below is part of the accepted answer

Abuse notation a tiny little bit to regard our injections as subset inclusion,and for any maximal ideal $m\subset B$ we have the situation $$ k\subseteq \frac{A}{\varphi^{-1}(m)}\subseteq \frac{B}{m}$$

So if $A$ and $B$ are $k$ algebras of finite type and $\varphi:A\longrightarrow B$ is homomorphism.For any maximal ideal $m\subset B$ we have

$f:k \rightarrow \frac{A}{\varphi^{-1}(m)}$ is injective & $g:\frac{A}{\varphi^{-1}(m)} \longrightarrow \frac{B}{m}$ is injective then Why do we have that subset inclusions? or could someone precisely explain that Abuse of Notation?

User
  • 15
  • If there is an injective map $f:A\to B$ that preserves whatever structure there is on $A$ and $B$, then you can identify $f(A)$ with $A$ and think of $A$ as 'sitting inside' $B$, in which case the injective map is regarded as an inclusion $A\hookrightarrow B$. Usually this would only be done when there is a 'natural' choice of injective map $f:A\to B$ though. If you think about how $\mathbb{Q}$ and $\mathbb{R}$ are constructed, the inclusions $\mathbb{Z}\hookrightarrow\mathbb{Q}\hookrightarrow\mathbb{R}$ are an abuse of notation in this manner also, but we lose nothing from it – user829347 Feb 09 '22 at 17:39
  • So any structure preserving injective map can be considered as an inclusion? – User Feb 09 '22 at 18:11
  • The point is that an injective homomorphism $A\to B$ is an isomorphism between $A$ and its image in $B$ (by Noether's first isomorphism theorem). – Douglas Molin Feb 09 '22 at 18:16
  • Yeah...now I feel it was trivial. Thank you! – User Feb 09 '22 at 18:41

0 Answers0