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Given a field $F$, I know that $F[X]$ is the ring of polynomials in $X$. I know that this is not a field.

I have seen the notation for $F(x)$ with round brackets. Usually when we use round brackets we take the smallest field containing $F$ and $x$. For example, if $F(\alpha) = F[\alpha]$ for all algebraic $\alpha \in E$ ($E$ some extension of $F$).

My question is what $F(x)$ is when $x$ is a variable. Is this just the quotient of polynomials? What is the definition? Is this what is called a function field.

John Doe
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2 Answers2

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$F(x)$ is:

  • The smallest field containing $F$ and the variable $x$;

  • The set of symbolic quotients $\frac{p(x)}{q(x)}$, where $p,q\in F[x]$ and $q\neq 0$.

  • The field of fractions associated to the ring $F[x]$.

Matthew Leingang
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Yes, you can define $F(X)$ as the quotient field of the ring of polynomials $F(X)$; informally, all rational expressions in $X$.

It is indeed an example of an (algebraic) function field in one variable (over $F$). But there are other (algebraic) function field in one variable (over $F$) too.

quid
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  • I think that $F(x)$ in the question is just a field of rational functions, i.e. quotient field of polynomials in one variable. Function fields are finite extensions of these. That is, when we add new elements to this field $F(x)$ and these elements generate a finite degree vector space over $F(x)$. – Mikhail D Aug 05 '19 at 21:46
  • I am not sure what you want to say. Yes, that's what I wrote. It is an example of function field but there are others too. – quid Aug 05 '19 at 21:52
  • I just opened Artin's Algebra for this question. He appears to define a function field as an extension of $\mathcal{C}(t)$, i.e. a function field must contain some extra elements. That is, he defines function fields as finite extensions of this specific quotient field. The number of variables has no relation to the definition. Still, I am in no way trying to challenge your answer. – Mikhail D Aug 05 '19 at 22:10
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    But $F(x)$ over $F(x)$ is an extension of degree $1$. Thus if Artin just says it's an extension I am pretty sure he'd not intend to exclude $F(x)$ it self. Maybe there is a more specific remark there; I'd be surprised though. – quid Aug 05 '19 at 22:15
  • Yes, $F(x)$ in this question is a trivial function field, but to make $F(x)$ a proper function field, based on Artin's Chapter 15, we have to add to it at least one extra algebraic element, not an extra variable. – Mikhail D Aug 05 '19 at 22:56