How to show that the homomorphism $\rho : \Gamma(V)_f \to F(D(f),k)$ is injective?
问题内容
I am self learning Algebraic Geometry from Daniel Perrin's Algebraic Geometry textbook.
I have a question on last paragraph of page $41$.
Let $D(f)$ be the set of points where the function doesn't vanish and $\Gamma(V)= k[X_1,...,X_n]/I(V)$ where $k$ is a commutative field.
Let $r$ denote the restriction homomorphism $r: \Gamma(V)\to F(D(f),k)$, where $F(D(f),k)$ denotes the ring of all functions from $D(f)$ to $k$. I donot understand why $r(f)$ is invertible?
$r(f)$ can be factorized through the localization $\Gamma(V)_f, r=\rho j$ and the homomorphism $\rho : \Gamma(V)_f \to F(D(f),k)$ is injective.
Author writes some explanation on why homomorphism in injective but I am not able to understand it: Indeed if $\rho(g/f^n)=0$, then $g(x)=0$ on $D(f)$ and hence $fg=0$ on $V$ , which implies that $g/f^n$ is zero in the localized ring.
How does $\rho(g/f^m)$ implies that $g(x)=0$ on $D(f)$? $g(x)=0$ on $D(f)$ implies that $fg=0$ but how does it implies that $g/f^n$ is zero in the localized ring?
Can you please help me with this?
Thank you very much!
回答 (1)
I think there are really only two points to check.
First, $r(f)$ is invertible on $D(f)$ because $f$ never vanishes there. Indeed, if $x\in D(f)$, then $f(x)\neq0$, so $1/f(x)$ is well-defined. Thus $r(f)$ has inverse $x\mapsto1/f(x)$.
Hence, by the universal property of localization, the restriction map factors through $\Gamma(V)_f$, with
$\rho\left(\frac{g}{f^m}\right)(x)=\frac{g(x)}{f(x)^m}.$
Now suppose $\rho(g/f^m)=0$. Then for every $x\in D(f)$,
$\frac{g(x)}{f(x)^m}=0.$
Since $f(x)\neq0$, it follows that $g(x)=0$ on $D(f)$.
From this, $fg$ vanishes on all of $V$: on $D(f)$ we have $g=0$, while outside $D(f)$ we have $f=0$. Therefore $fg=0$ in $\Gamma(V)$.
So in the localization,
$\frac{g}{f^m}=\frac{fg}{f^{m+1}}=0.$
This is exactly the usual criterion for a fraction to be zero in a localization: some power of $f$ annihilates the numerator. Here already $fg=0$.
Therefore $\rho$ is injective.