Questions about the proof of the isomorphism $k[X] \to \mathcal{O}(X)$ in the book Linear algebraic groups by Springer.

Question

I am reading the book linear algebraic group by Springer. I have some questions on page 8.

Theorem 1.4.5 is: $\phi: k[X] \to \mathcal{O}(X)$ is an isomorphism.

(1) Line 8-9 of the proof of Theorem 1.4.5, it is said that we may assume that $h_x=a_x$ since $D(a_x)=D(a_x^{n_x})$. I don't know why we may assume that $h_x=a_x$.

(2) Line 14 of the proof of Theorem 1.4.5, it is said that the ideal generated by $h_1^2, \ldots, h_s^2$ is $k[X]$ since $D(h_i)$ cover $X$. I don't know why the ideal generated by $h_1^2, \ldots, h_s^2$ is $k[X]$.

Could you explain this more explicitly? Thank you very much.

Answer 1

Question 1: There's a bunch of "replacing _ with _" implicit in the passage. Let's rewrite it.

Let $f \in \mathcal{O}(X)$. For each $x$ there exists an open neighborhood $U_x$ of $x$ and $g_x, h_x \in k[X]$ such that $h_x$ does not vanish on $U_x$ and $$f(y) = \frac{g_x(y)}{h_x(y)}$$ for each $y \in U_x$. Replacing $U_x$ with a smaller open neighborhood if necessary, we may assume that $U_x = D(a_x)$ for some $a_x \in k[X]$. Then $D(a_x) \subseteq D(h_x)$, so we have $$a_x^{n_x} = h_x h'_x$$ for some $h'_x \in k[X]$ and $n_x \geq 1$. So for $y \in U_x$ we have $$f(y) =\frac{g_x(y)}{h_x(y)} = \frac{g_x(y) h'_x(y)}{h_x(y) h'_x(y)}= \frac{g_x(y) h'_x(y)}{a_x(y)^{n_x}}.$$ Since $D(a_x) = D(a_x^{n_x})$, we may replace $a_x$ with $a_x^{n_x}$, so that $$f(y) = \frac{g_x(y) h'_x(y)}{a_x(y)}.$$ Now replace $g_x$ with $g_x h'_x$ and also replace $h_x$ with $a_x$, so that the above becomes $$f(y) = \frac{g_x(y)}{h_x(y)}$$ and we still have $U_x = D(a_x)$ and $h_x = a_x$.

Question 2: Well, if the $D(h_i)$ cover $X$, then the $D(h_i^2)$ also cover $X$, since $D(h_i) = D(h_i^2)$...

As an aside, my general feeling is that Springer's attempt to make the book totally self-contained by including all the necessary algebraic geometry is probably a bit too much for someone who doesn't actually know algebraic geometry yet. It's probably far easier to learn the relevant algebraic geometry elsewhere.