Why is $\mathbb{F}(x,y)$ not algebraic?

Let $\mathbb{F}$ be a field, why is $\mathbb{F}(x,y)$ not an algebric extension of $\mathbb{F}$? (Is $\mathbb{F}(x)$ an algebraic extension?)

*This is listed in my Field theory lecture notes as an important example for a finitely generated extension that is not algebraic and I don't know why it is not algebraic and if adding only one variable, $x$, is sufficient.

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1 Answer

Even $\mathbb{F}(x)$ (which is by definition the field of fractions of the polynomial ring $\mathbb{F}[x]$ in one variable) is not algebraic over $\mathbb{F}$. Namely, $x$ is an element which is transcendent by definition (read it).

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Moreover, extensions of the form $\mathbb{F}(x_1,x_2,\ldots,x_n)$ where the variables are assumed to have no relations are called free transcendental extensions, and it turns out that extensions (of finite transcendence degree, at least) can be decomposed into a tower of two extensions, one algebraic and one free. This means that understanding transcendental extensions only requires understanding algebraic extensions and understanding free extensions (sort of). –  Aaron May 22 '12 at 14:53