Let $x=x(T)$ and $y=y(T)$ be rational functions. Show that there is a non-constant polynomial $f(X,Y)$ such that $f(x,y)=0$? I am trying to solve the following problem:

*

*Let $x=x(T)$ and $y=y(T)$ be rational functions. Show that there is a non-constant polynomial $f(X,Y)$ such that $f(x,y)=0$.

And then, there is a "hint" (which I suspect is actually the entire answer due to its length):

*

*Let $K(T)$  be the field of rational functions with coefficients in the field $K$. If $x \in K(T)$ is non-constant, then $K(T)$ is an algebraic extension of the subfield $K(x)$ generated by $x$. In fact we have $x=\frac{p(T)}{q(T)}$ with $p,q$ polynomials, and then $T$ satisfies the polynomial equation $p(X)-x Q(x)=0$ and hence, every $y\in K(T)$ is algebraic over $K(x)$.

I am a bit confused:

*

*What is going on in this proof? I had a course in algebra with field extensions and such but due to the pandemic, the quality wasn't really good and I didn't learn it very well.

*Does this proof actually gives us methods to find a polynomial $f$ such that $f(x,y)=0$ or does it merely asserts that such polynomial must exist? There was a previous exercise to find a polynomial for some rational functions and I managed to answer it by trial and error but I couldn't use the content of this proof to obtain the desired polynomial.

 A: 
What is going on in this proof?

Field extensions can be confusing. That said, this is a rather vague question. I'll try to spell the argument out in detail. If nothing else, this might help you in pinpointing what it is that you're confused about.
Since $x \in K(T)$, we can find polynomials $f,g \in K[X]$ such that $x = \frac{f(T)}{g(T)}$. Rearranging this, we get the equation $f(T) - x \cdot g(T) = 0$. The left-hand side of this equation is the polynomial $f(X) - x \cdot g(X)$ evaluated at $T$. This is a polynomial (in the indeterminate $X$) with coefficients in the subfield $K(x)$ of $K(T)$. The fact that $T$ is a root of this polynomial means precisely that $T$ is algebraic over $K(x)$. Using basic results about field extensions* it follows that every element of $K(T)$ is algebraic over $K(x)$. In particular, there is a non-zero polynomial $F(X) \in K(x)[X]$ such that $F(y) = 0$.
Now $F(X)$ is a polynomial with coefficients in $K(x)$, so it has the form $\frac{f_n(x)}{g_n(x)}X^n + \ldots + \frac{f_1(x)}{g_1(x)}X + \frac{f_0(x)}{g_0(x)}$ for some polynomials $f_0,\ldots,f_n,g_0,\ldots,g_n \in K[Y]$, where $Y$ is a new indeterminate. This means that we have $G(x,y) = 0$ where $G(X,Y) = \frac{f_n(Y)}{g_n(Y)}X^n + \ldots + \frac{f_1(Y)}{g_1(Y)}X + \frac{f_0(Y)}{g_0(Y)}$. This is almost what we wanted to show - the only problem is that $G(X,Y)$ is not in $K[X,Y]$, but rather in $K(Y)[X]$. But that is easy to fix: we just clear the denominators by multiplying $G(X,Y)$ by $g_n(Y)\cdot g_{n-1}(Y) \cdot \ldots \cdot g_0(Y)$. The result will finally be a polynomial in the indeterminates $X$ and $Y$ that is satisfied by $x,y$.
Using slightly more high-tech language, what we have shown here is the not too surprising result that the transcendence degree of $K(T)$ over $K$ is $1$.

Does this proof actually gives us methods to find a polynomial $f$ such that $f(x,y)=0$ or does it merely asserts that such polynomial must exist?

It only gives existence. The problem is with the "using basic results about field extensions" part. This shows that $y$ is algebraic over $K(x)$ without explicitly giving us the polynomial $F(X)$ that is satisfied by $y$.
*I.e. the result that if $F \subseteq E$ are fields and $y \in E$, then $y$ is algebraic over $F$ if and only if $F(y)$ has finite dimension as a vector space over $F$.
