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What is algebraic function theory? In other words, what is an algebraic function field?

I need to understand the gist of it. Could someone explain what this is? I am writing a paper on the beginnings of ring theory.

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I'm not sure that "the gist" of a branch of mathematical theory is within the scope of this site. But what do I know ??? –  The Chaz 2.0 Mar 27 '12 at 4:21

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An algebraic function field is the pair consisting of a field $k$ (not assumed algebraically closed) and and a finitely generated extension field $K$ of transcendence degree one: $k\subset K=k(x_1,...,x_r), \; \operatorname {trdeg}_k K=1$.

There is an equivalence of categories between smooth irreducible projective curves $C$ over $k$ and function fields $K$ over $k$, in which $K$ corresponds to the field of rational functions on $C$.

Nowadays theoretical algebraic geometers tend to prefer the language of curves seen as schemes, whereas more applied mathematicians (and computer scientists) interested in the theory of codes and in cryptography seem to like function fields more.
It is well known that Grothendieck didn't like valuations, the core tool in function fields, and the dichotomy I evoke might partly be due to the tremendous influence of his and Dieudonné's treatise EGA.

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I've never heard the term, "algebraic function theory." I've heard of algebraic functions; a function $f(x)$ is algebraic if there is a polynomial $p(t)$ such that $p(f(x))$ is identically zero. Is that what you are talking about?

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Yes. I am reading Israel's Kleiner's "A History of Abstract Algebra" and this is exactly what he's talking about. I am trying to understand what were Emmy Noether's contribution to this field. She wrote a paper in 1927 entitled "Abstract development of ideal theory in algebraic number fields and function fields" where she talks about the decomposition of ideals as unique products of prime ideals. –  organgina Mar 27 '12 at 4:28
    
He uses "algebraic function theory" interchangably with "algebraic function field." Is that more clear? I need to know what an "algebraic function field" is all about. –  organgina Mar 27 '12 at 4:30
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Let $F$ be a field. I believe that algebraic function field is to $F(t)$ what algebraic number field is to the rationals, that is, it's a finite extension. –  Gerry Myerson Mar 27 '12 at 5:35

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