A local ring is defined as a ring which has a unique maximal ideal. This unique maximal ideal consists of only non-units and contains all the non-units of the ring $R$. So examples of local rings include any field, or rings localised at prime ideals and so on.

My question is: can we have a local ring which has more than one prime ideal? I can't seem to think of any examples where this is the case.

(In response to the comments below, I have realised that when $R$ is a domain, then $(0)$ is also a prime ideal, but are there any examples where $R$ isn't a domain?)

  • 1
    $\begingroup$ What about $\{0\}$? $\endgroup$ – Daniel Fischer Apr 30 '15 at 11:35
  • 1
    $\begingroup$ I would have thought 'most' examples one get to know have more than one prime ideal. For instance any commutative local integral domain which is not a field. $\endgroup$ – Jack Yoon Apr 30 '15 at 11:35
  • $\begingroup$ ${0}$ is a prime ideal if and only if $R$ is a domain isn't it, i was looking for examples where $R$ isn't necessarily a domain; maybe i should edit it in the question $\endgroup$ – user1314 Apr 30 '15 at 11:36
  • 1
    $\begingroup$ Polynomial ring would be one of a good example. Let $R$ be any commutative ring which is not a field, Then $R[x]$ has prime ideal $(x)$ which is not maximal. Maybe look at en.wikipedia.org/wiki/Krull_dimension if you want to know more about this. $\endgroup$ – Jack Yoon Apr 30 '15 at 11:41
  • 2
    $\begingroup$ It can have infinitely many, think to $\Bbb Q[x,y]_{(x,y)}$ and $(x+ay)$ for any $a \in \Bbb Q$. $\endgroup$ – Watson Feb 8 '17 at 19:07

This question is intimately tied to the idea of dimension. Given a commutative ring $R$, its (Krull) dimension is defined as the supremum of the lengths of chains of all prime ideals.

The name dimension for this quantity makes sense because the coordinate ring of an $n$-dimensional affine variety has Krull dimension $n$. Localization has a geometric interpretation, too: a prime ideal $\mathfrak{p}$ corresponds to an irreducible variety $V$. The localization $R_\mathfrak{p}$ consists of all rational functions that are defined at all points of $V$.

If $R$ is a local ring and its unique maximal ideal $\mathfrak{m}$ is the only prime ideal, then $R$ has dimension $0$, which, if $R$ is Noetherian, is equivalent to being Artinian.

The geometric interpretation above indicates how to produce a local ring with any number of prime ideals. Given a positive integer $n$, consider $n$-dimensional affine space $\mathbb{A}^n$ over a field $k$ of characteristic zero. Its coordinate ring is $k[x_1, \ldots, x_n]$, and if we localize at the prime (maximal, actually) ideal $(x_1, \ldots, x_n)$, we obtain the local ring $k[x_1, \ldots, x_n]_{(x_1, \ldots, x_n)}$ which consists of all rational functions on $\mathbb{A}^n$ that are defined at the origin. By the correspondence theorem for prime ideals in a localization, this ring has a chain of prime ideals $$ (0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \ldots \subsetneq (x_1, x_2, \ldots, x_n) $$ of length $n$.

As for an example where $R$ is not a domain: an affine variety is irreducible iff its coordinate ring is a domain, so we need to consider reducible varieties. Let $A = \frac{k[x,y]}{(xy)}$ which is the coordinate ring of the union of the lines $x = 0$ and $y = 0$ in the plane. Let $\mathfrak{m} = (x,y)$ and let $$ R = A_\mathfrak{m} = \left(\frac{k[x,y]}{(xy)}\right)_{(x,y)} $$ which is the local ring at the origin. Then $(x), (y)$, and $(x,y)$ are all prime ideals of $R$.


For an explicit example, consider $k[[x,y]]$, formal power series over a field $k$, in two indeterminates. The unique maximal ideal is the set of series with no constant term; among the nonzero prime ideals are $(x)$ and $(y)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.