I want to see why the following is true:
A Variety described by the Weierstrass equation has dimension 1.
Let $K$ be a field. An elliptic curve over $K$ is defined by the set of solutions in $\mathbb{P}^2(K)$ of a homogeneous Weierstrass equation $$E: Y^2 Z + a_1 XYZ + a_3 YZ^2 = X^3 + a_2 X^2 Z + a_4 XZ^2 + a_6 Z^3$$ with $a_1,a_2,a_3,a_4,a_5,a_6\in K$. And we claim $E$ to be non-singular.
As definition of the dimension:
Let $V$ be an affine (projective) variety. We define the dimension $dim(V)$ as the biggest length $n$ of a chain $S_0\supset S_1 \supset \dots \supset S_n$ of distinct irreducible closed subspaces of $V$.
In Galbraith book Exercise 5.6.5 says:
Let $f$ be a non-constant polynomial and let $X=V(f)$ be a variety in $\mathbb{A}^n$. Show that $dim(X)=n-1$.
A proof for this is given in Hartshorne's "Algebraic Geometry" Proposition 1.13. But I don't understand all the algebra used for the proof:
Proof: If $f$ is an irreducible polynomial, we have already seen that $Z(f)$ is a variety. Its ideal is the prime ideal $\mathfrak{p}=(f)$. By "Krull's Hauptidealsatz" $\mathfrak{p}$ has height 1, so by 1.8A, $Z(f)$ has dimension $n-1$.
(1.8A: Let $k$ be a field, and let $B$ be an integral domain which is a finitely generated $k$-algebra. Then:
(a) the dimension of $B$ is equal to the transcendence degree of the quotient field $K(B)$ of $B$ over $k$;
(b)For any prime ideal $\mathfrak{p}$ in $B$ we have $$ \operatorname{height}\mathfrak{p}+ \operatorname{dim} B/\mathfrak{p} = \operatorname{dim} B $$)
I don't understand the part (b) of the theorem 1.8A, what does the equation mean? What is $\operatorname{dim} B/\mathfrak{p}$, what does this notation mean?. Maybe there is a different approach to proof this Proposition?
I saw hints for this proof where they explain:
Let $f\in k[x_1,x_2,\dots,x_n].$ Without loss of generality suppose $f$ contains at least one monomial featuring $x_n$. Then write $f=f_r x_{n}^{r}+f_{r-1}x_{n}^{r-1}+\dots + f_0$ with $f_i\in k[x_1,\dots,x_{n-1}]$ or $k[x_0,\dots,x_{n-1}]$. One can check that the field $k(X)$ contains a subfield isomorphic to $k(x_1,\dots, x_{n-1})$. [$k(X)$ field of fractions of $k[X]$ over $k$, where $k[X]=k[x_1,\dots,x_n]/I_k(X)$] Finally, $k(X)$ is an algebraic extension of $k(x_1,\dots,x_n)$.
I would be really happy if someone could explain to me a little more detailed one of these proofs, or tell me another possible approach to prove this proposition.