3
$\begingroup$

I am reading Liu's Algebraic Geometry and Arithmetic Curves and get stuck at Lemma 8.1.2:

Let $A$ be a Noetherian ring an define for an ideal $I \subset A$ the $A$-algebra $$\tilde{A}:=\bigoplus_{d\geqslant 0} I^d, \qquad I^0:=A.$$ Let $f_1, \ldots, f_r$ be a system of generators of $I$ and let $t_i \in I = \tilde{A}_1$ denote the element $f_i$ considered as a homogeneous element of degree 1. We have a surjective homomorphism of graded $A$-algebras $$\phi: A[T_1, \ldots, T_n] \longrightarrow \tilde{A}, \qquad T_i \mapsto t_i.$$ Let $\tilde{X}:= \operatorname{Proj} \tilde{A}$.

My questions are the following:

  1. He says: If $P$ is a polynomial with coefficients in $A$, then $P(t_1, \ldots, t_n)=0$ if and only if $P(f_1, \ldots, f_n)=0$. Why is this worth a remark, if $f_i=t_i$?

  2. If $I$ is generated by a regular element, then $\tilde{A} \cong A[T]$. In the proof he says the homomorphism $\phi: A[T] \longrightarrow \tilde{A}$ from above is an isomorphism. But why? Is see, that the elements in $A[T]$ and in $\tilde{A}$ look very similar, but I cannot see, why the fact "regular" gives the claim.

I hope anybody can help me!

Thanks for helping me!

$\endgroup$
2
$\begingroup$
  1. The author says homogeneous polynomial. I think he wants to emphasize on the fact that $P(t_1,\dots,t_n)$ is a homogeneous element in $\bar A$.

  2. Let $f\in A[T]$, $f=a_0+a_1T+\cdots+a_nT^n$, such that $\phi(f)=0$. Suppose $(a)=I$. Then $f(a)=0$, that is, $a_0+a_1a+\cdots+a_na^n$. In $\bar A$ this leads to $a_ia^i=0$ for all $i$, so $a_i=0$ for all $i$ and thus $f=0$.

$\endgroup$
2
  • $\begingroup$ 1. Yes, i totally forgot, I am sorry. Okay thanks, that's fine. 2. Okay, and where do you actually use that $a$ is regular? At the point where you conclude $a_i=0$ from $a_ia^i=0$? $\endgroup$ – Arthur Dec 1 '15 at 9:10
  • $\begingroup$ Okay thanks I got ist. It's since $\mathcal{O}_{X,a}$ is integral... $\endgroup$ – Arthur Dec 2 '15 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.