Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From Mumford's Red Book, Chapter 2, Example K:

Take $X = Y = \mathbb{P}^2$, and let $x_0, x_1, x_2$ and $y_0, y_1, y_2$ be homogeneous coordinates on $X$ and $Y$. Let $U_0 \subset X$ and $V_0 \subset Y$ be defined as the open sets $x_0 x_1 x_2 \neq 0$ and $y_0 y_1 y_2 \neq 0$. Define an isomorphism between $U_0$ and $V_0$ by the map $y_i = 1/ x_i$. In fact, this is just an extension of the isomorphism of function fields:

$\begin{align} k\left(\frac{x_1}{x_0}, \frac{x_2}{x_0}\right) &\tilde\rightarrow k\left(\frac{y_1}{y_0}, \frac{y_2}{y_0}\right) \\\\ x_1/x_0 &\mapsto y_0/y_1 \\\\ x_2/x_0 &\mapsto y_0/y_2 \end{align}$

I'm confused as to what is meant by "an extension" here -- is this not just the map induced by this map of function fields? What am I missing?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

If $X$ and $Y$ are two varieties, then any isomorphism $K(X) \cong K(Y)$ arises from an isomorphism $U \cong V$ of non-empty open subsets $U$ and $V$ of $X$ and $Y$, but it need not extend to an isomorphism $X \cong Y$ (although if it does, the extension is unique). In this case it does so extend, and Mumford is remarking on that fact.

(The $X$ and $Y$ of my discussion are Mumford's $U_0$ and $V_0$; sorry for the notational clash.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.