8

Let $X,Y$ be varieties and suppose we have points $P \in X, Q \in Y$ such that the corresponding local rings are isomorphic, i.e. $\mathcal{O}_{Q,Y} \cong \mathcal{O}_{P,X}$. Then the problem is to show that there exist open sets $P \in U \subset X, Q \in V \subset Y$ and an isomorphism $U \cong V$ that takes $P$ to $Q$.

The solution given in http://math.berkeley.edu/~reb/courses/256A/1.4.pdf proceeds as follows: Let $f:\mathcal{O}_{Q,Y} \rightarrow \mathcal{O}_{P,X}$ be the isomorphism of local rings. We can consider $X,Y$ to be affine varieties of the same affine space $\mathbb{A}^n$. Moreover, we can take $P=Q=0$. Let $y_1,\dots,y_n$ be the coordinate functions on $\mathbb{A}^n$. Since $y_i$ is a regular function on $Y$, then it can be viewed as an element of the local ring $\mathcal{O}_{Q,Y}$, say $\langle V_i, y_i \rangle$. Under $f$ it is taken to a germ $\langle U_i, f(y_i) \rangle$. Define the open set $U$ of $X$ by $U = \cap_i U_i$. Then we can define a morphism $f^*: U \rightarrow Y$ by sending $(a_1,\dots,a_n) \in X$ to $(f(y_1)(a_1,\dots,a_n),\dots,f(y_n)(a_1,\dots,a_n))$. The solution then says that likewise we can define a morphism $g^*:V \rightarrow X$ on an open set $V$ of $Y$ and the composition of the two morphisms is the identity wherever it is defined.

Question 1: How exactly is $g^*$ defined? If i define it in the same manner as $f^*$ then i don't see why i get the identity.

Question 2: Why can we even compose $g^*$ and $f^*$? It seems that we have no guarantee that these morphisms are dominant.

Edit Any alternative solution to this problem will be accepted as answer as well. Edit

Bill
  • 439
Manos
  • 25,833
  • 1
    I don't have an alternative solution. Maybe look at Corollary 4.5 and work with the coordianate rings? Now that you know that the localization are isomorphic, then so will their quotient fields be (so $k(X)\cong k(Y)$). – quantum Mar 01 '19 at 14:36
  • 1
    A small comment: The isomorphism of local rings maps the only maximal ideals respectively to each other (It is an isomorphism and maps units to units) and from this we can see that we map $P$ to $Q$. Maximal ideal of $A_{\mathfrak{m}p}$ is of the form $\mathfrak{m}_p A{\mathfrak{m}_p}$ where $A=k[x_1,\cdots,x_n]/I(X)$ where $I(X)$ is prime and $I(X)\subset \mathfrak{m}_P$ –  Jul 31 '21 at 20:17

2 Answers2

3

Question 1: Let us represent a point of $\mathbb{A}^n$ as the action of an $n$-tuple of regular functions on the point itself. These regular functions are just the coordinate functions on $\mathbb{A}^n$. So we have that $a=(a_1,\dots,a_n) = (y_1,\dots,y_n)(a_1,\dots,a_n)$. Alternatively, the point $a$ is the image of itself under the identity morphism $(y_1,\dots,y_n)$ of $\mathbb{A}^n$. Now we can define a morphism $f^*: U \rightarrow Y$ by having $f$ act on the $n$-tuple of regular functions $(y_1,\dots,y_n)$ as $(a_1,\dots,a_n) \mapsto (f(y_1),\dots,f(y_n))(a_1,\dots,a_n)$, where $U=\cap_i U_i$ and $U_i$ is the open set where $f(y_i)$ is defined. For $g=f^{-1}$ let $V_i$ be the open set where $g(y_i)$ is defined and let $V=\cap_i V_i$. This gives a morphism $g^*: V \rightarrow X$. To see what is the effect of $g^*$ on the point $(f(y_1)(a),\dots,f(y_n)(a))$, observe that $(f(y_1),\dots,f(y_n))(a)$ and so by definition $f^*$ is induced by the action of $g$ on the $n$-tuple of regular functions $(f(y_1),\dots,f(y_n))$ giving $g^*((f(y_1),\dots,f(y_n))(a)) = (g(f(y_1)),\dots,g(f(y_n)))(a) = (y_1,\dots,y_n)(a)=(a)$.

Question 2: In general $f^*(U) \cap V$ will not be an open set. To remedy this, replace $U$ by $U^* := (f^*)^{-1}(V) \cap U$ and $V$ by $V^* :=(g^*)^{-1}(U^*) \cap V$. Then $f^*, g^*$ induce an isomorphism of $U^*,V^*$.

Manos
  • 25,833
  • I hope you agree with the comment: "What a mess!" – quantum Mar 01 '19 at 14:36
  • Thank you for your answer and @quantum for the comments. These help me a lot. Yet there is still one point not quite clear for me. Since in the solution manual quoted in the question, we have assumed $P=Q=0$ in $\mathbb{A}^n$. But I haven't spotted where you used this assumption that $P=Q=0$. Could you explain more on that? – Hetong Xu Jul 29 '20 at 05:12
1

Define $g^*$ in parallel to the way you define $f^*$, where you take $g = f^{-1}$ (which exists since $f$ is an isomorphism). Thus their composition is the identity since $f \circ g = g \circ f = id$.

As for dominance, we define the open sets $U$ and $V$ so that this is the case - i.e. define $U$ as above, then define $V$ as $f^*(U)\cap \cap_i V_i$ (where the $V_i$ are defined analogously to the $U_i$ above).

Dorebell
  • 4,154