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Let $R=\mathbb{C}[x]$ and $P=\{f(x)\in R|f(0)=0\}$. Then is $R_P$ a field?

I think so... my reasoning:

There's a bijective correspondence between prime ideals in $R_P$ and those in $R$ not intersecting $R\setminus P$ (Eisenbud Prop 2.2b). $R$ is a PID so every prime ideal is maximal. That is, prime ideals are of the form $(f(x))$ for some irreducible $f(x)\in R$. But $R\setminus P$ is the set of all functions with nonzero constant coefficient ($f(0)\neq 0$). Hence, if $(f(x))\cap (R\setminus P)=\emptyset$, then in particular $f(x)$ must have been in $R\setminus P$. But then $x|f(x)$ so $f(x)$ isn't irreducible. Therefore, there is no nonzero prime ideal that doesn't interset $R\setminus P$.

Hence, under the bijective correspondence, there are no nontrivial prime ideals in $R_P$. But since every nontrivial ideal can be extended to a maximal (prime) ideal, this implies that $R_P$ only has trivial ideals, i.e., $R_P$ is a field.

FTem
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  • would you not be able to show that P satisfies the field axioms? –  Mar 30 '17 at 20:50
  • I don't think I understand your question. Are you saying that $R_P$ is a field? – FTem Mar 30 '17 at 20:52
  • What is $R_P$? $\ $ –  Mar 30 '17 at 20:53
  • Where this goes wrong is in your conclusion that $x$ divides $f$. $x$ precisely divides those polynomials in $P$, not in its complement. – Tobias Kildetoft Mar 30 '17 at 20:54
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    @TobiasKildetoft $f$ is the generator of $P$ - so $f(x) = x$. In fact, that's where the proof goes wrong: "$x|f$, so $f$ is not irreducible". Not so, $f$ could just be equal to $x$. And it is. – Jack M Mar 30 '17 at 20:57
  • @JackM Ahh, I was not thinking this through. So his mistake is where you said, and he also has an error in the conclusion that $f$ is in $R\setminus P$ since the conclusion is precisely that it is in $P$. – Tobias Kildetoft Mar 30 '17 at 20:59
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    I think this is very relevant: http://math.stackexchange.com/questions/150892/when-the-localization-of-a-ring-is-a-field – DonAntonio Mar 30 '17 at 21:19

2 Answers2

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In a PID every nonzero prime ideal is maximal. But the zero ideal is prime, because the ring is a domain.

Since $P$ is a prime ideal (easy proof), there is a bijection between the prime ideals in $R_P$ and the prime ideals of $R$ contained in $P$. So $R_P$ has two prime ideals, hence it's not a field.

If $R$ is a domain, the only prime ideal $P$ such that $R_P$ is a field, is $P=\{0\}$. Indeed, the same argument as before applies: a nonzero prime ideal contains at least two prime ideals, itself and $\{0\}$.

egreg
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But then $x|f(x)$ so $f(x)$ isn't irreducible

This step is incorrect. If $x$ is a proper factor of $f(x)$ (that is, $x\mid f(x)$ and $f(x)/x$ is not a unit), then this would mean $f(x)$ is not irreducible. But $f(x)$ might be a unit times $x$. In fact, in that case $f(x)$ is irreducible, and $(f(x))=P$. So there is exactly one nonzero prime ideal in $R$ that does not intersect $R\setminus P$, namely $P$ itself. As a result, the localization $R_P$ has one nonzero prime ideal and is not a field.

Eric Wofsey
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