This is a proposition in Liu's book.(proposition 4.1.5)
I can't understand the proof….
Proposition 1.5. Let $X$ be an irreducible scheme. The following properties are equivalent:
(i) The scheme $X$ is normal.
(ii) For every open subset $U$ of$X$, $O_X(U)$ is a normal integral domain.
Proof: Let us suppose $X$ is normal. Let $U$ be an open subset of $X$ and $α ∈ Frac(O_X(U))$ integral over $O_X(U)$. Let $V$ be an affine open subset of $U$. Then $α$ is integral over $A := O_X(V)$ and $α ∈ Frac(A)$. Let us set $I = Ann((αA+A)/A)$. If $I \neq A$, then $I ⊆ p$ for a prime ideal $p$ of $A$.As $A_p$ is normal, there exists an $s ∈ A \setminus p$ such that $sα ∈ A$. It follows that $s ∈ I$, whence a contradiction. Hence $1 ∈ I$ and $α ∈O_X(V)$. As this is true for every affine open subset $V$ of $U$,we have $α ∈O_X(U)$ (Proposition 2.4.18). This shows that (i) implies (ii).
In proof why $\alpha$ is integral over $A := O_X(V)$ and $\alpha ∈ Frac(A)$
And why is $I$ defined like that?
I really need help with this proof….
Best Answer
Here by abuse of notation, Liu is identifying $\alpha\in Frac(O_X(U))$ with its image under the restriction map $O_X(U)\to O_X(V)$ which induces a map $Frac(O_X(U))\to Frac(O_X(V))$. So, this restriction of $\alpha$ is an element of $Frac(O_X(V))$ by definition. Since $\alpha$ is integral over $O_X(U)$, it is a root of some monic polynomial with coefficients in $O_X(U)$; applying the restriction map to the coefficients of this polynomial, we conclude that the restriction of $\alpha$ is a root of a monic polynomial with coefficients in $O_X(V)$, i.e. it is integral over $O_X(V)$.
The ideal $I$ is just the ideal of possible denominators of the fraction $\alpha$: that is, it is the set of elements $s\in A$ such that $\alpha=\frac{t}{s}$ for some $t\in A$. This is equivalently the set of $s\in A$ such that $s\alpha\in A$, which is equivalent to saying that $s\alpha A\subseteq A$ or that $s$ annihilates $(\alpha A+A)/A$ (which is the somewhat opaque definition of $I$ that Liu uses). So, if you can prove that $1\in I$, that means that $s$ can be written with a denominator of $1$, so it is just an element of $A$. So to prove this, you suppose that $1\not\in I$, so $I$ is a proper ideal that is contained in some prime ideal, and then you use the normality of the localization at that prime ideal to reach a contradiction.