I am trying to prove that the Krull dimension of an irreducible affine curve in $\mathbb{A}^2$ is 1. So my idea is a prove that the dimension of the coordinate ring is 1. So if the coordinate ring is
$$R = F[X,Y]/(f)$$ where $(f)$ is the irreducible polynomial in $F[X,Y]$ whose zero locus is the points on our irreducible affine curve. Now to show this has Krull dim 1, I was going to use prove that the transcendence degree of $R$ over $F$ is 1. I broke this into two cases:
Case 1: f is a polynomial of a single variable (say $Y$), then I am not sure if the following is correct but I think I can say that then $R \simeq F[X]$, and thus the transcendence degree of $R$ over $F$ is 1.
Case 2: f is a polynomial in two variables. Then I am thinking that I can say the transcendence degree of R over F is 1, if $[X] \in R$ is algebraically independent over $F$. This I am having trouble with, I think the idea is to prove that the homeomorphism $F[X] \to R$ that maps $X \to [X]$ is injective? And we also need to show that the fraction field $Frac(R)$ is algebraic over $F(X)$. Which I guess follows from the fact that $F$ is an algebraically closed field?
I'm wondering if somebody can point me in the right direction, or perhaps there is another way to look at this problem? I am not that comfortable with transcendence degree.