About Theorem 3.4 Hartshorne: detailed proof.

Question

I propose a detailed version of part of the proof of Theorem 3.14 from Hartshorne's book Algebraic Geometry. The questions are inserted from time to time within the proof. Thanks for your patience.

Notation Let $Y\subseteq\mathbb{P}^n$ be a projective variety. Letting $S=k[x_0,\dots, x_n]$ (equipped with the usual grading $S=\oplus_{r\ge 0} S_r$), there is a naturale grading on $S(Y)=S/I(Y)$ which is constructed as follows: for each integer $r\ge 0$ define $$S(Y)_r:=S_r/I(Y)_r:=\{f+I(Y)_r\;:\; f\in S_r\}$$ where $I(Y)_r=I(Y)\cap S_r.$

Theorem. Let $Y\subseteq\mathbb{P}^n$ be a projective variety with homogeneous coordinate ring $S(Y)$. Then we have:

$(a)\quad$ for any point $P\in Y$, let $\mathcal{m}_P\subseteq S(Y)$ denote the ideal generated by the set of all homogeneous elements $f\in S(Y)$ such that $f(P)=0$. Then $\mathcal{O}_P=S(Y)_{(\mathcal{m}_P)}$;

$(b)\quad$ $K(Y)\cong S(Y)_{((0))}$

Proof. Let $U_i\subseteq\mathbb{P}^n$ be the open set defined by $x_i\ne 0$ and set $Y_i:= Y\cap U_i$. We may consider $Y_i$ as an affine variety. We can construct a natural isomorphism $$\varphi_i^*\colon A(Y_i)\to S(Y)_{(x_i)}$$ as $$\varphi_i^*\bigg(f(y_1,\dots, y_{i-i},1, y_i, \dots, y_n)+I(Y_i)\bigg)=\frac{f(x_0,\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r},$$ where $f\in S_r$.

$(b)\quad$ Let $P\in Y$ and chose any $i$ so that the $i-$th coordinate of $P$ is non-zero. Then $P\in Y\cap U_i$. We may write $P=[a_0:\dots: a_{i-1}:1:a_i:\dots: a_n]$. Set $P_0:=(a_0,\dots, a_{i-1}, a_{i+1},\dots, a_n)=\varphi_i(P)$, the corresponding point in the affine space.

Denote with $\mathcal{m'}_{P_0}$ the maximal ideal of $A(Y_i)$ corresponding to $P_0$. More precisely, $$\mathcal{m'}_{P_0}=\bigg\{g+I(Y_i)\in A(Y_i)\;:\; g(P_0)=0\bigg\}.$$ It's easy to prove that: $$\boxed{\varphi_i^*(\mathcal{m'}_{P_0})=\mathcal{m}_P\cdot S(Y)_{(x_i)}}$$ Recall that the product in the right of the above equation is interpreted in the localization as follows (with the natural induced grading on the quotient ring $S(Y)=S/I(Y))$ $$\mathcal{m}_p\cdot S(Y)_{(x_i)}=\bigg\{\frac{f(x_0\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r}\;:\; f+I(Y)_r\in\mathcal{m}_P\cap S(Y)_r\;\text{for same}\; r\ge 0\bigg\}.$$

Using this isomorphism, we then have the isomorphism of the localizations $$A(Y_i)_{\mathcal{m'}_{P_0}}\cong \big(S(Y)_{(x_i)}\big)_{\mathcal{m}_P\cdot S(Y)_{(x_i)}}$$

Question 1. I believe that the last isomorphism can be obtained from exercise number 4 on page 44 of Introduction to Commutative Algebra by Atiyay - MacDonald, is that so?

We need to argue that $$\big(S(Y)_{(x_i)}\big)_{\mathcal{m}_P\cdot S(Y)_{(x_i)}}\cong S(Y)_{(\mathcal{m}_P)}\tag1$$

Question2. Why $(1)$ is not usual transitivity of localizations?

A typical element in $\big(S(Y)_{(x_i)}\big)_{\mathcal{m}_P\cdot S(Y)_{(x_i)}}$ is of the form

$$\frac{\frac{f+I(Y)_{s+r}}{x_i^{r+s}+I(Y)_{s+r}}}{\frac{g+I(Y)_r}{x_i^r+I(Y)_r}}\equiv \frac{f+I(Y)_{s+r}}{x_i^sg+I(Y)_{s+r}}$$ where $g+I(Y)_r\notin\mathcal{m}_p\cap S(Y)_r$, and the identification takes place in the quotient field of $S(Y)$. Now since $x_i^s+I(Y)_s\notin \mathcal{m}_p\cap S(Y)_s$ it follows that the product $x_i^sg +I(Y)_{s+r}\notin\mathcal{m}_p\cap S(Y)_{s+r}$. So the element on the right is a degree $0$ element of the localization

$$S(Y)_{\mathcal{m}_P}=\bigoplus_{n\in\mathbb{Z}}\big(S(Y)_{\mathcal{m}_P}\big)_n$$ where

$$\big(S(Y)_{\mathcal{m}_P}\big)_n=\bigg\{\frac{p+I(Y)_{n+r}}{q+I(Y)_r}\;:\; q+I(Y)_r\notin \mathcal{m}_P\cap S(Y)_r\;\text{for same}\; r\ge 0\bigg\}.$$ This proves that $$A(Y_i)_{\mathcal{m'}_{P_0}}\cong (S(Y)_{\mathcal{m}_p})_0=S(Y)_{(\mathcal{m}_P)}$$

$(b)\quad$ Observe that $K(Y)\cong K(Y_i)$. The final part is to show that $$K(Y_i)\cong S(Y)_{((0))}.$$ We have already obtained the isomorphism $\varphi_i^*\colon A(Y_i)\to S(Y)_{(x_i)}$. Extending this isomorpgism to $\tilde{\varphi_i^{*}}$ on the quotient fields fo both the sides we get $$\tilde{\varphi_i^*}\bigg(\frac{f(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}{g(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}\bigg)=\frac{\frac{f(x_0,\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r}}{\frac{g(x_0,\dots, x_n)+I(Y)_s}{x_i^s+I(Y)_s}}=\frac{x_i^s f(x_0,\dots, x_n)+I(Y)_{r+s}}{x_i^rg(x_0, \dots, x_n)+I(Y)_{r+s}}$$

In the denominator of the last expression above we have $g(x_0,\dots, x_n)+I(Y)_s\ne 0$ So to make sense of this, we need $x_i^r+I(Y)_r\ne 0$ as well. Since $I(Y)$ is a homogeneous prime ideal, this is violated only if $x_i\in I(Y)$. I have proved that $x_i\in I(Y)$ is equivalent to $Y_i=\emptyset$, which is absurd according to our choise of $U_i$.

Finally, notice that the element in the image of $\tilde{\varphi_i^*}$ in above are elements of $S(Y)_{((0))}$.

Question 3 It remains to show that every element of $S(Y)_{((0))}$ is also an image of an element of $K(Y_i)$. How can I show this?

My Solution for question 3.

A typical element in $S(Y)_{((0))}$ is of the form $$\frac{f(x_0,\dots, x_n)+I(Y)_r}{g(x_0,\dots, x_n)+I(Y)_r},$$ where $f+I(Y)_r\in S(Y)_r$ and $g+I(Y)_r\notin S(Y)_r\cap (0)$. Let us consider the dehomogenized of $f$ and $g$: $$f(y_1,\dots, y_{i-1}, 1, y_i,\dots, y_n)\quad\text{and}\quad g(y_1,\dots, y_{i-1}, 1, y_i,\dots, y_n),$$ where $\deg f(y_1,\dots, y_{i-1}, 1, y_i,\dots, y_n)= \deg g(y_1,\dots, y_{i-1}, 1, y_i,\dots, y_n) = r$, results

$$\begin{split}\tilde{\varphi_i^*}\bigg(\frac{f(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}{g(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}\bigg)=&\frac{\frac{f(x_0,\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r}}{\frac{g(x_0,\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r}}\\ =\frac{f(x_0,\dots, x_n)+I(Y)_{r}}{g(x_0, \dots, x_n)+I(Y)_{r}}\end{split}$$

The above happens just in case $x_i\nmid f$ and $x_i \nmid g$.

Suppose now that $x_i\mid f(x_0,\dots, x_n)$ and that $x_i\nmid g(x_0,\dots, x_n)$

We denote by $h(x_0,\dots, x_n)$ the homogenized polynomial of $f(y_1,\dots, y_{i-1},1,\,y_i,\dots, y_n)$, then $h\in S_{\deg f}$, then

$$ \begin{split} \tilde{\varphi_i^*}\bigg(\frac{f(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}{g(y_1,\dots, y_{i-1},1, y_i, \dots, y_n)+I(Y_i)}\bigg)= \frac{\frac{h(x_0,\dots, x_n)+I(Y)_{\deg f}}{x_i^{\deg f}+I(Y)_{\deg f}}}{\frac{g(x_0,\dots, x_n)+I(Y)_r}{x_i^r+I(Y)_r}}=&\frac{x_i^{r-\deg f} h(x_0,\dots, x_n)+I(Y)_{r}}{g(x_0, \dots, x_n)+I(Y)_{r}}\\ =&\frac{f(x_0,\dots, x_n)+I(Y)_r}{g(x_0,\dots, x_n)+I(Y)_r} \end{split} $$

Answer 1

Question 1: You can directly decude this isomorphism using Corollary 3.2 in Atyiah, MacDonald:

In general, let $A$ be a ring and let $S$ be a multiplicatively closed subset of $A$. Then the localization $S^{-1}A$ and the homomorhpism $f:A\to S^{-1}A: x\mapsto x/1$ have the following properties:

1. $s\in S\Rightarrow f(s)$ is a unit in $S^{-1}A$.

2. $f(a)=0 \Rightarrow as=0$ for some $s\in S$.

3. Every element of $S^{-1}A$ is of the form $f(a)f(s)^{-1}$ for some $a\in A$ and some $s\in S$.

Further, Corollary 3.2 states that if $g:A\to B$ is a ring homomorhism with this three properties, then $B$ is isomorphic to $S^{-1}A$.

Now let $h:S(Y)_{(x_i)}\to \big( S(Y)_{(x_i)} \big)_{\mathfrak{m}_P\cdot S(Y)_{(x_i)}}: h(x)=x/1$ with the properties from above. Further, let $S=A-\mathfrak{m'_p}$. Then

$$ h\circ\varphi^*_i: A(Y_i) \to \big( S(Y)_{(x_i)} \big)_{\mathfrak{m}_P\cdot S(Y)_{(x_i)}}$$

fulfills the properties from above. A the proof is simple when using that $\varphi^*_i$ is an isomorphism and that $h$ fulfills these properties. Consequently, by the corollary,

$$ \big( S(Y)_{(x_i)} \big)_{\mathfrak{m}_P\cdot S(Y)_{(x_i)}} \cong A(Y_i)_{\mathfrak{m}'_P}.$$

Question 2:

This should precisely be the transitivity of the localization, see also this question.

Question 3:

Now our construction so far does not make use of the fact that $\mathfrak{m}_P$ is a maximal ideal. Indeed, it suffices that it is prime. Therefore, for question 3 we can just consider the prime ideal $(0)$ as $A(Y_i)$ is a domain. $\varphi^*_i$ maps this to $(0)\subseteq S(Y)$, thus localizing and taking the degree $0$ elements yields this isomorphism, since $K(Y_i)=A(Y_i)_{(0)}$.