Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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
2  
@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
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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