Decomposing field extentions into algebraic and purely trancendental extentions

Question

Given a field extension $K/k$ we can take the algebraic (integral) closure of $F / k$ in $K$ and then $F/K$ only has transcendental elements. However in this MO post the top answer talks about the 'reverse': where $K$ is an algebraic extension of some intermediate field that is purely transcendental over $k$. In the context of the post $k$ is the prime subfield, but I don't believe that should matter and the notion should apply an extension of any field.

The problem is I do not see how this is constructed. We cannot simply take the collection of all elements transcendental over $k$ as the field it generates will contain algebraic elements assuming $K$ has them to begin with. Actually as I was writting this I realized we can use AC to construct a chain $k < k(\gamma_0) < k(\gamma_0, \gamma_1) ...$ of purely trancendental extentions which must terminate and then $F = \lim k_{\alpha}$ is our desired field. Still I would like to know what the differences/advantages between these two constructions are and the situations in which they are each useful.

Answer 1

It seems to me that you are looking for the notions of transcendence degree and/or transcendence basis.

If $K/k$ is an extension of fields, then we say that a subset $B\subset K$ is a transcendence basis of $K/k$, if

• $B$ is algebraically independent over $k$, i.e. the elements of $B$ do not satisfy any non-trivial multivariable polynomial equation with coefficients from $k$, and
• The extension $K/k(B)$ is algebraic. In other words, every element of $K$ is algebraic over $k(B)$.

It turns out that all transcendence bases of a given extension $K/k$ have the same cardinality. But, there is nothing canonical about the intermediate field $k(B)$. In particular the degrees $[K:k(B)]$ can vary wildly when we vary the basis $B$.

For a specific example consider the function field $K=k(x,y)$ of an elliptic curve $$y^2=x^3+ax+b$$ defined over a field $k$ such that $6\cdot1_k\neq0_k$. Clearly $K$ is algebraic over either $k(x)$ or $k(y)$, so either $x$ or $y$ (they are both transcendental over $k$ by definition) will form a transcendence basis of $K/k$. So does any monomial $x^m$ or $y^n$. The transcendence degree $\operatorname{tr.deg}(K/k)=1$. But $[K:k(x)]=2$, $[K:k(y)]=3$, $[K:k(x^n)]=2n$ et cetera.

In this example case a nice geometric result is that the field $K/k$ is not purely transcendental. IOW there cannot be an element $z\in K$, transcendental over $k$, such that $K=k(z)$. This is possible only for function fields of curves of genus zero. As the elliptic curves have genus one, such an element $z$ cannot exist.