I've just learned about the idea of local ring, but the only concrete examples in the book are $\{p/q \in \mathbb{Q} \mid q \mbox{ odd}\}$ and the power series ring $k[[x]]$ for some field $k$. But the only prime ideals they have are just $\{0\}$ and their maximal ideal.

Can anybody provide me with more concrete examples of local rings with "non-trivial" prime ideals?


First consider the polynomial ring $R=F[x,y]$ for a field $F$. One chain of prime ideals in this ring is $(0)\subseteq (x)\subseteq (x,y)$. Now $(x,y)$ is a maximal ideal, but there are other maximal ideals, for example $(x+1,y)$.

The easiest way to eliminate the other maximal ideals is to pass to the localization at the prime $M=(x,y)$ so that the new ring $R_M$ is a local ring. It is a property of localization that the prime ideals contained in $M$ will have prime counterparts in $R_M$, with the same containment relations and everything. Thus the chain $(0)_M\subseteq (x)_M\subseteq (x,y)_M$ will be a properly ascending chain of prime ideals in the new ring $R_M$.

Even if you are not handy with localization now, you probably will need to be soon, so trying to understand this type of example is worthwhile.

  • $\begingroup$ Why is $(x+1,y)$ a maximal ideal? $\endgroup$ – inbrevi Jun 19 '20 at 15:18
  • $\begingroup$ When you quotient by it you get $F$. $\endgroup$ – rschwieb Jun 19 '20 at 15:35
  • $\begingroup$ Why? Could you explain it further? $\endgroup$ – inbrevi Jun 19 '20 at 15:40
  • $\begingroup$ @inbrevi $F[x,y]/(x+1,y)\cong F$. By correspondence of ideals in the quotient ring, $(x+1,y)$ is maximal. If that is tripping you up, I invite you to search for some posts explaining quotient rings. $\endgroup$ – rschwieb Jun 19 '20 at 16:23
  • $\begingroup$ Thanks! But somehow I just don't get this isomorphism. It is easy to see if $x+1$ is replaced by $x$. But why does it also hold for $x+1$? $\endgroup$ – inbrevi Jun 19 '20 at 17:53

If $k$ is any field then the ring of formal power series in two variables $k[[x,y]]$ is a local ring with "non-trivial" prime ideal $(y)$ since $k[[x,y]]/(y)\cong k[[x]]$.

More generally: $k[[x_1,...,x_n]]$ is a local ring with maximal ideal $(x_1,...,x_n)$ and prime ideals $(x_{i_1},...,x_{i_m}),\,1\le i_1 < \cdots i_m \le n$.


You will soon learn about localization and that will provide many many examples.

An example: Let $R$ be the ring of rational functions $f/g\in\mathbb Q(x,y)$ such that, when expressed in irreducible terms, the denominator is not divisible by $x$.

  • $\begingroup$ The OP wants a local ring of Krull dimension $>1$. $\endgroup$ – user89712 Dec 2 '13 at 17:20

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