I am reading Fulton's book Algebraic Curves. Currently I am working on a specific problem (2.43), and I have doubts about my work and would appreciate another opinion(s).

Assume $p$ is the origin in $\mathbb{A}^n$ and $\mathcal O_p(\mathbb{A}^n)$ is the set of all rational functions defined on $\mathbb A^n$ and $m_p (\mathbb{A} ^n)$ is the set of non units. Show $I\mathcal O_p = m_p$ so $I^r\mathcal O_p = m_p^r$ where $I$ is the ideal generated by $x_1,...,x_n$.

My proof seems too simple and that is what bothers me.

$(\supset)$ Let $\phi \in m_p$ thus $\phi = \frac{f}{g} $ such that $f(p)=0$. Well $I \mathcal O_p$ is the set generated by $\frac{r}{u}$ where $r \in I$. Thus because $f(p)=0$ this implies it is in $\mathcal I(V(p))$ thus $f \in I$

$(\subset)$ Let $\phi \in I\mathcal O_p$ thus $\phi= \frac{k}{h}$ where $k \in I =<x_1,x_2,..x_n>$ thus $\phi \in m_p(\mathbb{A}^n)$

I am not very sure on the second part, but how does this look?

  • 1
    $\begingroup$ Fulton's book is a tough one, I've already read it. Your proof seems correct to me. In the first part, remember only to use the fact that $P=0\in\mathbb{A}^n$. At the second part, remember to use the fact that, in any ring, any ideal is contained in a maximal ideal (and the inclusion follows because there's only one maximal ideal in the local ring). By the way, once you are about to start section $2.9$, prepare yourself: proposition $6$ was the most difficult one to me. $\endgroup$
    – Larara
    Feb 26 '16 at 7:20

The idea of the proof is correct. The proof is simple, but conveys the important relationship between $\mathcal{I}(p)$, the maximum ideal of polynomials vanishing on $p$, and $m_p$, the maximum ideal of rational functions vanishing on $p$. They are based on the same property of vanishing on $p$, which is not affected by denominators!

One important fact to state is that $\mathcal{I}(p) = I$. With this, the logical flow of the end of the first and second part of your proof is clear: $f(p) = 0$ implies $f\in\mathcal{I}(p) = I$ (note: not $\mathcal{I}(V(p))$!) for the first part, and $k\in I = \mathcal{I}(p)$ implies $k(p) = 0$ for the second.

In fact, the proof can be concisely written as follows:

$$ \phi= \frac{f}{g}\in m_p \Leftrightarrow \frac{f(p)}{g(p)} = 0 \Leftrightarrow f(p) = 0 \Leftrightarrow f\in \mathcal{I}(p) = I\Leftrightarrow \phi = \frac{f}{g}\in I\mathcal{O}_p $$


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