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Consider the following situation:

Let $k$ a field $X,Y$ indeterminates over $k$ and set $S=k[[X,Y]]/(Y-X^2)$ and $R=k[[y]]$, where $y$ is the class of $Y$ in $S$. Then both $S,R$ are regular but $S/yS = k[[X]]/(X^2)$ is not.

Question: Bruns and Herzog, CMR, bottom of page 68 say that the reader "should imagine the geometry" of this example. Could someone please explain to me as simply as possible what is this geometry?

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up vote 3 down vote accepted

Martin, please correct me if i'm wrong.

This is a supplement to Martin's correct answer.

Manos, maybe you're missing that the ring you mentioned is not the coordinate ring of the affine variety that you call the origin. The latter would be $k[X,Y]/(X,Y) \cong k$, whereas your ring is $k[X,Y]/(X^2,Y)$. Its underlying topological space is indeed a single point, however one can check that the local ring is not regular, as we do now:

Instead of $k[X,Y]/(X^2,Y)$ we continue with the isomorphic ring $R = k[x]/(x^2)$.

Some preliminaries. In $R$, the element $x$ is nilpotent, therefore any prime ideal must contain $x$. However, $R/(x)=k$ is a field, so $(x)$ is maximal. So $(x)$ is the only prime ideal and also the only maximal ideal.

Knowing this, we go on. The local ring at the origin (which is the only point anyway) is $R_{(x)}$, but since localizing to an ideal means formally inverting all elements outside of the ideal, and since all elements outside of $(x)$ are already units, this does nothing. So $R_{(x)} \cong R$.

The question is now: why is this not a regular local ring? Well, since there is only one single prime ideal, its Krull dimension is zero. However, the minimal amount of generators for the unique maximal ideal is clearly not zero, in fact it is one. So the ring is not regular.

To link back to Martin's comment, one should think of the intersection of the parabola and the $x$-axis as a "double point". This information can't be seen in the underlying topological space, but it can be seen in the defining equations (it's the $x^2=0$ part), the result is that this is a scheme with a single point, and this single point is not smooth! In more general terms we are dealing with a nonreduced scheme, whereas all varieties are reduced. So one has to let go a bit of classic geometric intuition.

Hope this helps.

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Dear Joachim, your answer is fantastic! – Manos Jan 31 '14 at 1:57
That's great to hear, @Manos. – Joachim Jan 31 '14 at 19:22

Let's forget about power series and work with polynomials instead.

$K[X,Y]/(Y-X^2)$ is the coordinate ring of the parabola, which is non-singular. In fact it is isomorphic to the affine line. Accordingly, we have a canonical isomorphism $K[X,Y]/(Y-X^2) \cong K[X]$.

However, $K[X,Y]/(Y-X^2,Y)$ is not regular, since this is the coordinate ring of the intersection of the parabola with the $X$-axis, which is not transversal.

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Good, what do you mean by "transversal"? – Manos Jan 28 '14 at 20:18
Can you give me a connection to the fact that "a regular local ring corresponds to a smooth point"? I see that $K[X,Y](Y-X^2,Y)$ is the coordinate ring of the origin $(0,0)$. Then? – Manos Jan 28 '14 at 20:33
You can find this in every text on alg geom or comm alg. – Martin Brandenburg Jan 28 '14 at 22:06

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