# If $(A/\mathfrak{m})_{\pi(f)}$ is a field then $A/\mathfrak{m}$ is a field

Let $k$ be a field and let $A$ be a finitely generated $k$-algebra, $\mathfrak{m} \subseteq A$ a prime ideal, and $\pi: A \to A/\mathfrak{m}$ the canonical projection. I'm trying to prove that for any $f \in A$, if the localization $(A/\mathfrak{m})_{\pi(f)}$ is a field then $A/\mathfrak{m}$ is a field, but I've been getting a bit confused.

$\textbf{My Attempt}$: We first note that there is an inclusion preserving bijection between prime ideals of $(A/\mathfrak{m})_{\pi(f)}$ and prime ideals of $A/\mathfrak{m}$ that do not contain $\pi(f)$. But $(A/\mathfrak{m})_{\pi(f)}$ is a field, and so the only prime ideal of $(A/\mathfrak{m})_{\pi(f)}$ is the zero ideal. Hence the only prime ideal of $A/\mathfrak{m}$ that does not contain $\pi(f)$ must also be the zero ideal.

Now suppose $\mathfrak{m}$ is not maximal. Then there exists a maximal ideal $\mathfrak{m} \subseteq \mathfrak{m'} \subseteq A$ with $f \in \mathfrak{m'}$. But $A$ is a finitely generated $k$-algebra, implying that $\mathfrak{m}$ is the intersection of all such ideals. Since $f$ is also in the intersection, we have that $f \in \mathfrak{m}$, a contradiction. Hence it must be that $\mathfrak{m}$ is maximal and $A/\mathfrak{m}$ a field.

Does this work? I'm very unsure of the proof as it is.

• Looks fine. One comment for clarity: I would say that $m$ is the intersection of all maximal ideals $m'$ containing $m$ before noting that each such ideal contains $f$. Your current wording suggests that only some of these maximal ideals contain $f$, and intersecting just those ones gives $m$. – stewbasic Sep 7 '16 at 5:58
• You can simplify your question by realizing that you actually want to prove the less crowded assertion: "if $R$ is an integral domain and for all $f\in R$ the localization $R_f$ is a field then $R$ is a field." – Hamed Sep 7 '16 at 6:05
• @Hamed Don't you at least need the assumption that $R$ is a finitely generated $k$-algebra? Otherwise wouldn't you be able to prove that the set of closed points in $SpecR$ is dense for any ring $R$? – leibnewtz Sep 7 '16 at 7:34
• Maybe some of you are thinking in "for all $f$" and some other in "for a single $f$"? – A.G Sep 7 '16 at 7:57

I think a good way to think about the problem is geometrically. Write $X=\operatorname{Spec}⁡R$ and $U=\operatorname{Spec}⁡R_f$.Then the inclusion $\pi : U \to X$ is a morphism of finite type. We already know that $U$ consists of a single point, say $U=\{u\}$. Since $\pi$ is dominant we must have $\pi(u) = \eta_X$, the generic point of $X$. On the other hand $\pi(u) = \eta_X$ is constructible by Chevalley's Theorem, in particular closed ($X$ is finite type over a field!) and so it follows that $X$ itself is a field.