Computing the dimension of a ring [duplicate]

Suppose that $k$ is an algebraically closed field, and consider $k[x,y]$, the ring of the polynomials in two variables over $k$. If I have an irreducible polynomial $p(x,y)$, then $R:=k[x,y]/p(x,y)$ is an integral domain. How can I compute it's dimension in general? If you want an example you can refer to the polynomial in my previous question (Show that $(x^2+y^2+a^2)^2-a^4-4a^2x^2$ is irreducible for $a\neq 0$), but just give me a general strategy or a hint, don't compute the dimension already please

Question: "How can I compute it's dimension in general?"

Answer: If $p$ is a non-constant irreducible polynomial in $A:=k[x,y]$, it follows the krull dimension equals $krdim(A/(p))=1$. You may find a proof of this result in a book on commutative algebra or algebraic geometry (Hartshorne, Prop.I.1.13).

@Carlos - as mentioned in the other answer/comment: It is called the "Krull principal ideal theorem" - a result from commutative algebra. Youll find this in mostbooks on commmutative algebra.

Comment: "I wouldn't use that proposition explicitly because I didn't saw it in my course, but maybe this proposition could work for the polynomial p(x,y) in the example: if A→B is an integral ring morphism, krdimB≤krdimA. Since p(x,y) can be regarded as a monic polynomial in k[y][x], it follows that the morphism of rings k[y]→R is integral, and so krdimR≤krdimk[y]=1. Since R is an integral domain (0) is prime and since (p) is not maximal (in k[x,y]), krdimR≥1 too. – Carlos 11 hours ago"

Response: In general if you are given a prime ideal $I:=(f_1,..,f_l) \subseteq A:=k[x_1,..,x_n]$, it can be difficult to calculate the Krull dimension of $B:=A/I$ dimension explicitly. You may find a result and some references in Hartshorne (Ch.I.Thm.1.8). It says that $krdim(B)=trdeg_k(K(B))$ where $K(B)$ is the quotient field of $B$. Here $trdeg_k(K(B))$ is the transcendence degree of the field $K(B)$ over $k$.

Example: If $B:=k[x,y]$ is the polynomial ring it follows $K(B)=k(x,y)$, and the two variables $x,y$ are "transcendent over $k$". Hence $trdeg_k(k(x,y))=2$. (See Matsumura, "Commutative ring theory").

Hence given a $k$-algebra $B$ which is a domain, you must construct algebraically independent variables $x_1,..,x_n \in K(B)$ (meaning the elements $x_i$ does not satisfy any polynomial relation over $k$) and then prove that $k(x_i)=K(B)$. From this it follows that $trdeg_k(K(B))=n=krdim(B)$.