While working on exercise 3.20 in Hartshorne I've gotten stuck on something for a while. Given an integral scheme $X$ of finite type over a field I'm tring to show the existence of some affine open subset $U$ which has $\dim U=\dim X$ (I know that all open subsets satisfy this according to part e but I haven't gotten this far yet). If I can prove this then I have completed the proof of $a$ and $b$.
I know from general topology that $\dim U\leq\dim \bar{U}=\dim X$. I'm not sure how to show the other inequality. I'm not sure if it's pure topology or if it involves some of the properties of a scheme, but for some reason it is just driving me crazy and I'm not seeing it.
I'd appreciate any help. Thanks.
I don't see a way to do this, without going basically the whole way to the proof. You know that $\dim X=\sup_\alpha \dim U_\alpha$ for any open cover $\{U_\alpha\}$ of $X$. That said, now each $U_\alpha=\text{Spec}(A_\alpha)$ is an integral affine $k$-variety of dimension $d_\alpha:=\dim(A_\alpha)$ (Krull dimension). Noether Normalization then gives you a finite map $U_\alpha\to \mathbb{A}^{d_\alpha}$. Clearly then
$$d_\alpha=\text{tr.deg}_k{k(x_1,\ldots,x_\alpha)}=\text{tr.deg}_k(\text{Frac}(A_\alpha))$$
since $\text{Frac}(A_\alpha)/k(x_1,\ldots,x_\alpha)$ is algebraic.
Since $\text{Frac}(A_\alpha)\cong \text{Frac}(A_\beta)$ as $k$-algebras for all $\alpha,\beta$, it follows that $\dim(U_\alpha)=\dim(U_\beta)$ for all $\alpha,\beta$, and in fact, it's equal to $\text{tr.deg}_k \mathcal{O}_{X,\eta}$, where $\eta$ is the generic point of $X$.
I think the key somehow is that it's not easier to show that some affine open has the same dimension as $X$, but to notice that all affine opens need to have the same dimension as each other.