ideas algunas

---------------------------------------------------------------------------------- Idea 1. Let $k$ be any field. Suppose that $x,y$ are elements such that $k(x),k(y)$ are isomorphic as $k$-algebras to the field of rational functions. If you adjoin the elements $x,y$ to the field $k$ and occurs that $k(x,y)/k(x)$ is a finite extension (i.e., $y$ is algebraic over $k(x)$), then there must be some property concerning to a field extension of $k(y)$ in order to prove that $x$ is algebraic over $k(y)$, and therefore, that $k(x,y)/k(y)$ is also finite. ---------------------------------------------------------------------------------- Idea 2. If you have a finite extension $k(x)/k$, then take the minimal polynomial of $x$ over $k$. Reciprocally, if you want an extension $k(x)/k$ to be finite, try to prove that $x$ is algebraic over $k$ (and make sure it actually happens over $k$). ---------------------------------------------------------------------------------- Example: Let $k$ be any field, and let $L/k(x)$ and $K/k(y)$ be two finite extensions. Assume that both $k(x)$ and $k(y)$ are isomorphic to the field of rational functions in one variable. Assume that $K\subseteq{L}$. Then $L/K$ is a finite extension.____________________________ Solution: We can try to apply our idea 1, since $L/k(x)$ is finite and $k(x)\subseteq{k(x,y)}\subseteq{L}$ both implies that $k(x,y)/k(x)$ is finite. We would like to show that $k(x,y)/k(y)$ is finite (no matter what is the problem asking us to prove), so following our idea 1, we must find some property that shows what we're looking for. We find it using our idea 2. Taking $f(Y)$ the minimal polynomial of $y$ over $k(x)$, we notice that since $y$ is algebraic over $k(x)$, then it follows that $x$ is algebraic over $k(y)$ (and therefore $k(x,y)/k(y)$ is finite) if and only if $f(Y)\notin{k[Y]}$ (otherwise we couldn't get a polynomial in terms of a variable $X$ with coefficients over $k(y)$ where $y$ vanishes). Since $f(Y)\notin{k[Y]}$ because $y$ is not algebraic over $k$, we're done.