Let $k=\bar k$ be a field. I'm trying to "write down" the residue fields at various points on $\mathbb{A}^n=\operatorname{Spec} k[x_1,\cdots, x_n]$, but am having some trouble with the non-closed points. The definition I'm using is that residue field at a point $x$ on an integral scheme $X$ is the residue field of the local ring $\mathcal{O}_x$. Thus, if $\mathfrak{p}$ is a point on the affine scheme $X=\operatorname{Spec} A$ then the residue field at $\mathfrak{p}$ is $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$, which is isomorphic to the fraction field of $A/\mathfrak{p}$.
On $\mathbb{A}^1=\operatorname{Spec} k[x]$ these fields are not hard to determine. The only points are the generic point $(0)$ and the closed points $(x-a)$ for $a\in k$. Clearly the residue field at the generic point (i.e., the function field) is $k(x)$ and the residue fields at closed points are all just $k$.
Already for $\mathbb{A}^2$, though, I'm finding this a bit trickier. The points in this case are again $(0)$ (generic), $(x-a,y-b)$ for $a,b\in k$ (closed), and $(f)$ for $f\in k[x,y]$ irreducible (nonclosed). The function field at the generic point is clearly $k(x,y)$, and it's also not hard to see that the function fields at closed points are all $k$. But what do we get for the function field at a nonclosed point $(f)$? I think this should come out to be an algebraic extension of a transcendence degree 1 extension of $k$, but would like some clarification on this. My thoughts are as follows. Given $f\in k[x,y]$ irreducible, the function field, which we denote $k(f)$, will be the fraction field of $k[x,y]/(f)$. Setting $R=k[x]$ and $g(y)=f(x,y)\in R[y]$, we have $k[x,y]/(f)=R[y]/(g(y))$, which I believe is isomorphic to $R(\alpha)$, where $\alpha\in \overline{k(x)}$ satisfies $g(\alpha)=0$, i.e., is a root of a polynomial equation with coefficients in $k[x]$. Thus, the fraction field of $k[x,y]/(f)$ should be an algebraic extension of $k(x)$. Morally, this feels right to me in that the various function fields one obtains for $\mathbb A^2$ include transcendence degree 0,1,2 extensions of $k$. What makes me somewhat skeptical is the choice $R=k[x]$ was completely arbitrary. The same reasoning would also show that $k(f)$ is an algebraic extension of $k(y)$. In either case, $k(f)$ seems to be a trans. degree 1 extension of $k$, but the degrees $[k(f):k(x)]$ and $[k(f):k(y)]$ need not be equal, which I find a bit disturbing. Can someone tell me what's going on here?
In the general case $\mathbb{A}^n$, what I expect then is that the function fields are again $k$ at the generic point, $k(x_1,\cdots, x_n)$ at the closed points, and ... transcendence degree $n-1$ extensions of $k$ at the nonclosed other points?
Let's take an example. Consider an elliptic curve, for argument's sake $$y^2=x^3+x$$ over $k=\Bbb C$. The residue field at its generic point is the fraction field of $$\frac{\Bbb C[x,y]}{(y^2-x^3-x)}.$$ This is $\Bbb C(x)[y]$ where $y$ is a square root of $x^3+x$, so a quadratic extension of $\Bbb C(x)$ and also it is $\Bbb C(y)[x]$ where $x$ is a root of the equation $x^3+x-y^2=0$. Thus the function field is a degree $3$ extension of $\Bbb C(y)$.
This behaviour is quite typical. If you like consider the rational curve $y^2=x^3$: its function field is $\Bbb C(t)$ where $t=y/x$, $y=t^3$ and $x=t^2$. Then $|\Bbb C(t):\Bbb C(x)|=|\Bbb C(t):\Bbb C(t^2)|=2$ and $|\Bbb C(t):\Bbb C(y)|=|\Bbb C(t):\Bbb C(t^3)|=3$.