1

In the proof of affine dimension theorem (you can see the proof I am following here), Hartshorne assume that $A(Y)/\mathfrak{p}$ is the coordinate ring of the irreducible component $W$.

Why is this true?

qwerty89
  • 210
  • 1
    Look at the previous sentence. Minimal primes lying over $(f)$ are in bijection with irreducible components, so modding out by such a prime yields the coordinate ring of such a component, and Hartshorne just assigns the name $W$ to that component. – Tabes Bridges Apr 03 '21 at 01:21
  • Yes, but by the definition of coordinate ring, for $W$ must be $A(W)=k[x_1,...,x_n]/\mathfrak{p}$, and I don't know why this is the same of $A(Y)/\mathfrak{p}=(k[x_1,...,x_n]/(f))/\mathfrak{p}$. – qwerty89 Apr 03 '21 at 08:12
  • 1
    Since $(f) \subset \mathfrak p$, we can complete $f$ to a generating set $f, g_1,...,g_j$ of $\mathfrak p$. Both rings are the result of setting all $j+1$ generators equal to $0$ in $k[x_1,...,x_n]$; the only difference is that one of the descriptions does this in two steps. Does that clarify anything? – Tabes Bridges Apr 04 '21 at 01:26

1 Answers1

0

Ok let me try to explain. We have that some variety $W$$\hookrightarrow$$\mathbb A^n_k$ is irreducible by assumption. So to answer your question, we really just have to prove that $W$=$Z(p)$ is irreducible iff $p$ is a prime ideal in $k[x_1,...,x_n]$.

Assume that $W$=$Z(p)$=$X_1 \cup X_2$=$Z(p_1) \cup Z(p_2)$. We have by proposition $1.1.1$ in Hartshorne, $W$=$Z(p_1p_2)$, so we have applying ideals on both sides that $p$=$p_1p_2$, a contradiction. (Prime ideals are irreducible).

Armed with this result, we can now apply the coordinate ring functor $A$ and obtain our result. In other words, $A(W)$=$A(\mathbb A^n_k)$/$I(W)$=$A(\mathbb A^n_k)$/$p$.

combinatorialist46Carey2
  • 501