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Consider the domain $R=\mathbb{C}[x,y]/(y^2-x^3)$.

What would be an example of a chain of prime ideals of $R$ of maximal length?

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up vote 4 down vote accepted

The hardest part here seems to be proving that $y^2 - x^3$ is irreducible, which I'll attempt to show below. In general, an irreducible hypersurface in $\mathbf A^n$ has dimension $n - 1$. This is proved in most texts on algebraic geometry — see, for example, Proposition 2.25 of Milne's notes. [Prof Emerton notes in the comments that here $R$ is visibly finite over $\mathbf C[x]$, and hence these two rings have the same dimension.] So take the zero ideal of $R$ and a maximal ideal (corresponding to a point on the curve) of $R$.

To conclude, suppose we have $y^2 - x^3 = f(x, y)g(x, y)$, where $y$ occurs in $f(x, y)$. Suppose that $\deg_yf = 1$. Then $\deg_y g = 1$ and we can write $$ f(x, y) = a(x)y + b(x) \qquad \text{and} \qquad g(x, y) = c(x)y + d(x). $$ Then $a(x)c(x) = 1$, so after shuffling around constant factors we can have $a = c = 1$. Multiplying things out, we get $$ y^2 - x^3 = y^2 + y(b(x) + d(x)) + b(x)d(x). $$ So we must have $d = -b$ and hence $\deg b = \deg d$. But then the degree of the left side with respect to $x$ is even, which is absurd. The case of $\deg_yf = 2$ can be treated using the same ideas and is slightly easier to boot.

I would be very interested in a more geometric or at least conceptual way of looking at this. My dimension-fu is a little rusty.

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Please write about why $y^2-x^3$ is irreducible. Thanks! – Dave Sep 11 '11 at 19:13
Sure. I've been building furniture and trying to think of a less ad hoc way of doing this, but I'll just write up the first thing that came to mind and we'll see if it's right :) – Dylan Moreland Sep 11 '11 at 22:07
Dear Dylan, You don't need the hauptidealsatz. If you observe that $\mathbb C[x,y]/(y^2 - x^3)$ is finite over $\mathbb C[x]$, it follows that it has dimension $1$. Best wishes, – Matt E Sep 12 '11 at 5:10
@Matt That's a very good point. It is sad to see all of the German words leave this post, but so it goes. – Dylan Moreland Sep 12 '11 at 5:23
What is meant by $R$ is finite over $\mathbb{C}[x]$? – Dave Sep 12 '11 at 20:29

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