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I'm reading the book of Silverman about elliptic curves. It describes the function field of a variety defined over K to be the quotient field of K[X]/I(V), then says that we may look at the function field of P^n (projective n-space) as the subfield of Kbar(X_0...X_n) consisting of rational functions F(X)=f(X)/g(X) for which f,g are homogeneous with the same degree. How do those two definitions agree, and how are such functions a subfield of Kbar(X_0...X_n)?

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2 Answers 2

Where Silverman defines the function field of $V$ as the fraction field of the intergral domain $K[X]/I(V)$ he is talking about an integral affine variety $V$. In this case, $I(V)$ is a prime ideal so that $K[X]/I(V)$ is a domain.

If you look at Silverman's definition of function field for a projective integral variety $V$ (see page 10 in the Arithmetic of Elliptic Curves), you'll note that he defines it as the function field of $V \cap \mathbb{A}^n$ where $\mathbb{A}^n$ is one of the standard affine patches of $\mathbb{P}^n$ that intersects $V$. The (minor) subtlety is that more than one standard affine patch might intersect $V$ so you might think you would get different function fields by picking different affine patches. But this doesn't happen, and the remark you mention (remark I.2.9) describes the function field canonically as a subfield of $K(X_0,\ldots,X_n)$. The equivalence of the two definitions then follows immediately from the definition of the different standard open affine patches of $\mathbb{P}^n$ (i.e. de-homogenizing by dividing by a non-zero homogeneous coordinate).

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The function field of $V$ is defined as the field of fractions of $K[X]/I(V)$ for affine varieties $V$. In the case of projective varieties, Silverman chooses a Zariski-dense affine open subset $V$ of the variety and defines the function field of the variety as the function field of the subset $V$. Of course, one can prove it is independent of the choice of $V$. When the variety is $\mathbb{P}^n$, choose $V = \mathbb{A}^n$ and so $I(V)=\{ 0 \}$, so $K(\mathbb{P}^n) = K(V) = K[X]/\{ 0 \} = K[X]$, where $X$ is shorthand for $X_1,\ldots,X_n$. Finally to get an isomorphism of the subfield of $K(X_0,\ldots,X_n)$ you described with the field I just described, just set $X_0=1$.

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