Commutative ring with an ideal that contains all the nonunits

Is there an example of a commutative ring with an ideal that contains all the non-units?

I was trying to think of some subring of $\mathbb Q$, but I couldn't get it to work.

• Great question! Look up Local Ring. – Ravi Jun 12 '14 at 5:00
• Technically $\mathbb Q$ is such a ring. A rather boring example; every field is a local ring. – Dustan Levenstein Jun 12 '14 at 6:55
• The ring $\mathbb Z_{(2)}\subset \mathbb Q$ consisting of fractions with odd denominator is an example of what you require. – Georges Elencwajg Jun 12 '14 at 9:33
• Yes, and the fractions with even numerators is the ideal of nonunits. – Nishant Jun 12 '14 at 16:11
• Yes. Georges' example is one of what is called a "Discrete Valuation Ring", which is a special kind of local ring with the property that the nonzero proper ideals of $R$ are precisely the powers $\mathfrak m^k$ for $k \ge 1$, where $\mathfrak m \subset R$ is the maximal ideal. Consider the local ring $R/\mathfrak m^2$. – Dustan Levenstein Jun 13 '14 at 0:28

Atiyah-Macdonald's book: "Introduction to commutative algebra" explains what Jake says:

Proposition 1.6. i) Let $A$ be a ring and $\mathfrak{m} \neq (1)$ an ideal of $A$ such that every $x \in A - \mathfrak{m}$ is a unit in $A$. Then $A$ is a local ring and $\mathfrak{m}$ is its maximum ideal.
.... ii) Let $A$ be a ring and $\mathfrak{m}$ a maximal ideal of $A$, such that every element of $1 + \mathfrak{m}$ (i.e., every $1 + x$, where $x \in \mathfrak{m}$) is a unit in $A$. Then $A$ is a local ring.

Proof. i) Every ideal $\neq (1)$ consists on non-units, hence is contained in $\mathfrak{m}$. Hence $\mathfrak{m}$ is the only maximum ideal of $A$.

So note that for a commutative ring $R$ these are equivalent:

1. The ring $R$ is local ,
2. The set $m$ of non-units is an ideal,
3. The set $m$ of non-units is the only maximal ideal.

For example the ring $k[[X]]$, where $k$ is a field.

Take the ring $R$ of $2 \times 2$ matrices $A = [a_{ij}]$ with $a_{11} = a_{22}$ and $a_{21} = 0$ over a field $F,$ and let $I$ be the ideal consisting of those matrices in $R$ with with $a_{11} = a_{22} = 0.$

• This is correct. The ring $R$ is isomorphic to the ring of dual numbers $F[X]/(X^2)$. We commutative algebraists can smell in Geoff a member of the rival gang of non-commutative algebraists :-) Anyway, +1 – Georges Elencwajg Jun 12 '14 at 9:51
• @Georges: Yes, upper triangular matrices with constant main diagonal entries are my natural choice of local ring, but usually non-commutative! – Geoff Robinson Jun 12 '14 at 12:56

Take $R=\mathbb Z / 9 \mathbb Z$ and $I=3R$. Then $I=\{0, 3, 6\}$, which are exactly the non-units.

More generally, $R=\mathbb Z / p^2 \mathbb Z$ and $I=pR$, where $p$ is a prime.