# Chain of prime ideals of maximal length

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?

-

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.
Please write about why $y^2-x^3$ is irreducible. Thanks! –  Dave Sep 11 '11 at 19:13
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
What is meant by $R$ is finite over $\mathbb{C}[x]$? –  Dave Sep 12 '11 at 20:29