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Wikipedia says that a commutative ring $A$ is a field iff $A[x]$ is a PID.

The "only if" part is easy: we just apply the Euclidean algorithm. I've stumbled trying to prove the "if" part, though.

$\newcommand{\aa}{\mathfrak{a}}$ My best attempt so far: suppose $\aa$ is an ideal of $A$, and $\aa'$ is the ideal of $A[x]$ spanned by $i(\aa)$, where $i: A \hookrightarrow A[x]$ is the natural embedding. $i^{-1}(\aa') = \aa$ for grading reasons ($i(A) = A[x]^0$, and multiplication by a non-zero degree polynomial takes us out of $i(A)$, because $A \cong i(A)$ is integral). $A[x]$ is a PID so $\aa' = (a')$, where $a' = i(a)$ for some $a \in A$. Therefore, $\aa = (a)$, and thus $A$ is also a PID.

That's the best way to use the fact that $A[x]$ is a PID that I've found so far. Now I feel like there must be a trick to show that if $a \neq 0$ then $a = 1$, but I don't know how to do this. Any hints?

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    $\begingroup$ Consider the ideal (x) in $A[x]$. $\endgroup$ – Sungjin Kim Apr 10 '13 at 8:56
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Consider the surjective map $A[x]\to A[x]/(x)\cong A$. By the Lattice (or Fourth) Isomorphism Theorem the ideals of $A$ are in bijection with those in $A[x]$ containing $(x)$. Since $A[x]$ is a PID, any ideal containing $(x)$ must be of the form $(f)$ for some $f\mid x$. Thus either $f$ is a unit multiple of $x$ or it is a unit, so $(f)=(x)$ or $(f)=(1)$. Thus the only two ideals of $A$ are $(0)$ and $(1)$, so it is a field.

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I am sure that this is more complicated than necessary.

Note that since $A[x]$ is a domain, so is $A$, and we can use the identity $$ \deg(a b) = \deg(a) + \deg(b)\tag{degree} $$ for $0 \ne a, b \in A[x]$.

Take any $a \in A$, $a \ne 0$, and consider the ideal $(a, x)$ of $A[x]$.

This is principal, so there is $c \in A[x]$ such that $(a, x) = (c)$, and thus $a = b c$ and $x = d c$ for some $b, d \in A[x]$. By (degree) we have that $c \in A$, and thus $d = u + v x$, for $u, v \in A$. Thus $x = (u + v x) c = u c + v c x$, so that $c$ is invertibile and $(a, x) = (c) = A[x]$.

In particular, there are $s, t \in A[x]$ such that $a s + x t = 1$. (This is because $(a, x) = \{ a s + x t : s, t \in A[x] \}$.) Set $x = 0$ to find that $a$ is invertible in $A$.

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  • $\begingroup$ why "Set $x=0$"? $as+xt=1\implies (as,xt)=(1,0)\implies as=1$ $\endgroup$ – zed111 Apr 27 '15 at 16:07
  • $\begingroup$ Not necessarily, for instance it could be $a = t = 1$, $s = 1 - x$. $\endgroup$ – Andreas Caranti Apr 27 '15 at 17:29
  • $\begingroup$ Oh yes I missed $s,t \in A[x]$ $\endgroup$ – zed111 Apr 27 '15 at 17:35
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$1=\dim(A[x]) \geq \dim(A)+1$ gives $\dim(A)=0$, and $A$ is an integral domain, thus $A$ is a field.

The same proof a little bit more detailed: If $A$ is not a field, choose a prime ideal $\mathfrak{p} \neq 0$, then $0 \subseteq \mathfrak{p} \subseteq \mathfrak{p}[x]$ is a proper chain of prime ideals of $A[x]$. In a PID every prime ideal $\neq 0$ is maximal. Hence $A[x]$ is not a PID.

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  • $\begingroup$ +1. A proof that avoids considering the irreducibility of the polynomial $x$. I'm not sure the latter holds without assumption that the base ring is a domain, so I was wondering if in your proof one really needs the hypothesis that $A[X]$ is a domain. In other words isn't it true that any nonzero prime ideal of a principal ideal ring is maximal? I think I could prove this, but I'd have to use the Zariski-Samuel theorem; yet there may be something more elementary. $\endgroup$ – Marc van Leeuwen Apr 14 '13 at 11:15
  • $\begingroup$ Hm, isn't the question equivalent to if prime elements are irreducible? The usual proof only works for integral domains. I don't know for the general case. Maybe you can ask this as a separate question so that it gets more attention? $\endgroup$ – Martin Brandenburg Apr 14 '13 at 11:40
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    $\begingroup$ Tanks for the suggestion, which in fact had crossed my mind. See math.stackexchange.com/questions/361258/…. $\endgroup$ – Marc van Leeuwen Apr 14 '13 at 13:00
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    $\begingroup$ Using AG should be considered cheating - too goddamn beautiful 8) $\endgroup$ – Alexei Averchenko Apr 14 '13 at 18:01
  • $\begingroup$ This is only CA (beyond element calculations). But you are right, AG is beautiful! $\endgroup$ – Martin Brandenburg Apr 14 '13 at 19:32

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