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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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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!2011-09-11
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    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 :)2011-09-11
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    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,2011-09-12
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    @Matt That's a very good point. It is sad to see all of the German words leave this post, but so it goes.2011-09-12
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    What is meant by $R$ is finite over $\mathbb{C}[x]$?2011-09-12
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    @Robert Just that $R$ is finitely generated over $\mathbf C[x]$ as a module -- the relation allows us to write any power of $y$ as a $\mathbf{C}[x]$-linear combination of $\{1, y\}$.2011-09-12
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    The upshot for dimensions is that the field extension corresponding to the inclusion $\mathbf C[x] \subset R$ will be finite, so the transcendence degree (which is one way to define the dimension in this case) won't go up.2011-09-12
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    You could also argue that a finite extension of rings is in particular integral, and it follows from the going-up theorem that the dimensions don't grow under an integral extension. See pg 38 of [Hochster's notes](http://www.math.lsa.umich.edu/~hochster/614F10/614.pdf).2011-09-12