Let $\varphi:A\rightarrow A'$ be an integral ring extension.

1) Show that for every maximal ideal $m'\subset A'$ the ideal $\varphi^{-1}(m')\subset A$ is maximal.

2) and that for every maximal ideal $m\subset A$ there is a maximal ideal $m'\subset A'$ with $\varphi^{-1}(m')=m$.

Now there is a tip given: for $m\subset A$ look at $S=A-m$ and the localization $S^{-1}A$ and $S^{-1}A'$. Now I didn't get any further with this tip! Please help me!

1) What I did: $\varphi^{-1}(m')$ is a prime ideal (else $m'$ wouldn't be one in $A'$ as well, so also not maximal). Now I define the map $\psi:A/\varphi^{-1}(m')\rightarrow A'/m',a+\varphi^{-1}(m')\mapsto a+m$. Now $\psi$ is well defined as if $a-b\in \varphi^{-1}(m')$ then $\psi(a-b)=0\iff\psi(a)=\psi(b)$. So now if $x\in A'$ then there are $a_i\in A$ s.t. $x^n+a_1x^{n-1}+\cdots+a_n=0$. But this is inducing $(x+m)^n+(a_1+m)(x+m)^{n-1}+...+(a_n+m)=m$, so $\psi$ is an integral ring extension of integral domains (as $\varphi^{-1}(m')$ is prime) with $A'/m'$ a field, so then $A/\varphi^{-1}(m')$ is also a field and thus $\varphi^{-1}(m')$ is maximal in $A$.

For 2) I didn't use that the extension is integral, so I'm not sure if it's correct at all: $m\subset A\Rightarrow \exists m'\subset A'$ s.t. $m\subset m'$. Now $\varphi^{-1}(m')=m$, because if $x\in\varphi^{-1}(m')-m$ then $\varphi^{-1}(m')$ is still an ideal, not containing $1$, as $m'$ is maximal, so its a proper ideal which contains $m$ in $A$. So now here the only detail I'm missing is in this step: $\exists m'\subset A'$ s.t. $m\subset m'$. But I figured I can look at the ideal $(m)$ in $A'$, which is the ideal generated by every element of $m$, so then there exists a maximal ideal which contains the ideal $(m)$ properly, which isn't possible. Now I would only need to show that $(m)$ is a proper ideal, but im not sure how, if its true at all in general, but I guess it should be...

So now my question is

a) how can I use the tip?

b) did I make any mistakes or not show something important?


Hint: (1). Show that if $\phi: A \to B$ be an integral extension, then for any multiplicative closed sets $S \subset A, S^{-1}A \to S^{-1}B$ is an integral extension.

(2). Take $S:= A - \mathfrak m.$ Let $\alpha: A \to S^{-1}A, \beta: B \to S^{-1}B, \gamma:S^{-1}A \to S^{-1}B $ be natural maps. Then $\beta \phi = \gamma \alpha.$ Let $\eta$ be a maximal ideal of $S^{-1}B.$ Show that $\mathfrak m = \phi^{-1}\beta^{-1}(\eta).$ Here $\beta^{-1}(\eta)$ is a prime ideal in $B.$

  • $\begingroup$ So after using Hint (1) and 1) I get $\alpha^{-1}(\gamma^{-1}(n))=\alpha^{-1}(mS^{-1}A)$. Now I don't know how I can argue that $\alpha^{-1}(mS^{-1}A)=m$ and $\beta^{-1}(n)$ is prime. I'm having trouble with possible zero divisors! Till now we only viewed integral domains... A prime ideal doesn't have to contain all zero divisors does it? It would only need to contain "half" of them? Do maximal ideals have to contain all of them? $\endgroup$ – user208995 Jan 19 '15 at 20:47
  • $\begingroup$ @user208995 I would suggest reviewing the basic properties of localization as given in, say, Prop 2.2 of Eisenbud's book. Preimages of primes are always primes (algebraic geometry would fall apart if this were not true). To answer the last question: no. $k \times \{0\}$ is a maximal ideal of $k \times k$, for example. $\endgroup$ – Hoot Jan 20 '15 at 1:53
  • $\begingroup$ Ok, thank you, I will look into it today afternoon! $\endgroup$ – user208995 Jan 20 '15 at 8:20
  • $\begingroup$ Ok, I've got it now: preimages of primes are primes, I can show this by using that if $R\rightarrow R'$ is an injection and $R'$ an integral domain, so is $R$, as I can view $R$ as a subring of $R'$. So now $\alpha^{-1}(mS^{-1}A):=d$ is prime and clearly $m\subset d$ but $m$ is maximal, so $m=d$. Now I know $\beta^{-1}(n)$ is prime in $A'$ so there is a maximal ideal containing it, which preimage under $\beta$ must be $m$ by similar idea as I've given in my proof attempt in the question. Did I understand everything right? $\endgroup$ – user208995 Jan 20 '15 at 12:24
  • 2
    $\begingroup$ As $n$ is maximal it's preimage under $\gamma$ has to be maximal, as shown, so it is the unique maximal ideal of $S^{-1}A$ $\endgroup$ – user208995 Jan 20 '15 at 14:20

The first part looks okay, modulo this fact about integral extensions of domains. I assume this is something you can just use. For the second part, we definitely need to use the integral hypothesis because $\mathfrak{m}A' \neq A'$ needn't be true otherwise: $\mathbf{Z} \subset \mathbf{Q}$ probably yields the simplest example.

The condition $\mathfrak{m}A' \neq A'$ ought to remind you of Nakayama's lemma; that's why we localize at $\mathfrak{m}$ and henceforth assume that $A$ is local. There is some work hiding here: After localization, why do I still have an integral extension? How does one transport the result for that extension back to the original problem? Having justified this, we would be done if we knew that $A'$ were a finitely generated $A$-module, but you can use the usual trick of an ideal being the unit ideal iff it contains $1$ in order to work with a ring that is a finitely generated $A$-module.

I'm happy to say more, but I wanted to give the main ideas and see how that went first.

  • $\begingroup$ I don't get the part with the $A$-Module $A'$. In which way would it help me with this problem, if I knew it would be finetly generated? $\endgroup$ – user208995 Jan 20 '15 at 12:42
  • $\begingroup$ @user208995 You want to rule out the possibility that $\mathfrak{m}A' = A'$. You know $A \neq 0$ because it has a prime ideal and you know $A \subset A'$, so $A'$ is non-zero. Nakayama is good for getting contradictions to stuff like this. $\endgroup$ – Hoot Jan 20 '15 at 14:14
  • $\begingroup$ The above is too much work given part (a), which I addressed and then totally forgot. Still, I think it's a nice trick so I'll leave it up. $\endgroup$ – Hoot Jan 20 '15 at 18:21

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