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Just Reviewing some elementary algebra and would just like to have my proof checked.

Show that the following conditions on a ring $R$ are equivalent:

  1. $R$ has only one maximal ideal.
  2. The nonunits of $R$ form an ideal.

Proof: Let $R$ be a ring with unique maximal ideal $\mathfrak{m}$ and let $f : R \to R/\mathfrak{m}$ be the quotient mapping. We note that $R/\mathfrak{m}$ is a field and therefore has two ideals, namely $\langle 0 \rangle$ and $\langle 1 \rangle$. For $r \in R$ not a unit, we observe that $r \in \ker f$, indeed, since ring homomorphism preserve units. Therefore the set of all non-units in $R$ lie in the kernel of $f$, which is an ideal. Moreover, this ideal does not contain any non-units, since otherwise, these would lie in the image of $f$.

For the converse, suppose the non-units form an ideal in $R$, call this ideal $I$. By definition, $I$ is contained in some maximal ideal $\mathfrak{m}$. Now suppose $R$ has two distinct maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$. We may then consider two quotient mappings $$f: R \to R/\mathfrak{m}, \ \ \text{and} \ \ g : R \to R/\mathfrak{n},$$ where $R/\mathfrak{m}$ and $R/ \mathfrak{n}$ are fields. Note that $\ker(f) = I = \ker(g)$, from the argument above. The first isomorphism theorem then gives us that $\mathfrak{m} = \mathfrak{n}$.

Note: The last step I included a commutative diagram, but Math.Stack does not seem to support tikz.

Cheers.

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  • $\begingroup$ Where in the proof of $1 \Rightarrow 2$ did you use the fact that $\mathfrak m$ is unique? $\endgroup$ – user491874 Dec 3 '17 at 23:53
  • $\begingroup$ Your $1\Rightarrow 2$ argument seems to be based on the incorrect notion that ring homomorphisms send non-units to non-units. Also, the sentence "this ideal does not contain any non-units, since otherwise, these would lie in the image of $f$" is confusing. First, I assume "non-units" is a typo for "units". More seriously, without qualification "the image of $f$" means $f[R]$ but $f$ is surjective so every element of $R/\mathfrak m$ is in $f[R]$. Perhaps the image you meant was $f[\mathfrak m]$, but in general the image of an ideal could contain a unit, so this argument is incomplete at best. $\endgroup$ – Derek Elkins Dec 4 '17 at 5:02
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$1\Rightarrow 2$: Let $\mathfrak m$ be the unique maximal ideal of $R$ and $x\in R$ is nonunit. The principal ideal $xR$ is then a proper ideal, and, as a consequence of a (presumably known, based on Zorn’s lemma) theorem it is contained in a maximal ideal of $R$. Because $\mathfrak m$ is unique, that maximal ideal must be $\mathfrak m$, thus $xR\subseteq \mathfrak m$ and therefore $x\in\mathfrak m$. This proves $\mathfrak m$ is the set of all nonunits, which is therefore an ideal.

$2\Rightarrow 1$: Let $\mathfrak m$ be the ideal made up of non-units. Any proper ideal $\mathfrak n$ can only consist of non-units, so $\mathfrak n\subseteq \mathfrak m$. If $\mathfrak n$ is maximal, this implies $\mathfrak n=\mathfrak m$, i.e. $\mathfrak m$ is the unique maximal ideal.

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