4
$\begingroup$

I'm trying to prove that $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$, where $\mathfrak{a}$ is an ideal of $A = K[x_1, ... , x_n]$ and $K$ is an algebraically closed field. In case this notation is nonstandard:

if $T \subseteq A$ then $Z(T) = \{P \in \mathbb A^n \mid f(P) = 0 \ \forall f \in T\;\}$

and if $Y \subseteq \mathbb A^n$, then $I(Y) = \{f \in A \mid f(P) = 0 \ \forall P \in Y\;\}$.

Using Hilbert's Nullstellensatz I can get that $ I(Z(\mathfrak{a})) \subseteq \sqrt{\mathfrak{a}}$. I'm having trouble with the other direction though. Suppose $f \in \sqrt{\mathfrak{a}}$. Then $f^r \in \mathfrak{a}$ for some integral $r > 0 $. I want to show that $f$ annihilates everything in $Z(\mathfrak{a})$, given that $f^r$ annihilates everything in $Z(\mathfrak{a})$. I can't see how to make it happen, though.

Any help would be great. Thanks

  • 0
    \frak{a} is $\frak{a}$2012-01-18
  • 0
    You might want the symbol $\mathfrak{a}$, which is a Fraktur letter 'a' (Fraktur letters are often used for ideals of rings, among other things). You can get Fraktur characters with the command \mathfrak.2012-01-18
  • 4
    **Note:** `\mathfrak` is the better command for this site; for example, `$\frak{a} \subseteq A$` produces $$\frak{a} \subseteq A$$ while `$\mathfrak{a} \subseteq A$` produces $$\mathfrak{a} \subseteq A$$ (i.e. `\frak` changes the current font to Fraktur, while `\mathfrak` applies it only to its argument). However one could use `${\frak a} \subseteq A$`, which produces $${\frak a} \subseteq A$$ (the curly braces contain the effects of the `\frak` command)2012-01-18
  • 0
    This is very useful, @Zev: I had always be bothered by the fact that \frak changes the current font and had to resort to clumsy curly braces to remedy that. So thanks a lot for solving my problems for the second time in a few minutes (and now, without my even having to ask!)2012-01-18
  • 0
    @Georges: Always happy to help! In truth, I'm not entirely sure *why* `\frak` displays this behavior on the site; it doesn't when I use LaTeX on my computer, for example. Since `\frak` [is officially deprecated](http://www.tug.org/texlive/Contents/live/texmf-dist/doc/fonts/amsfonts/amsfonts.pdf) (see page 6) in favor of `\mathfrak`, it throws a warning, but it acts like one would expect, i.e. how `\mathfrak` does. I suspect they changed the name since `\mathfrak` follows the pattern of `\mathcal` and `\mathbb`.2012-01-18
  • 0
    By the way, to format in-line code the way I have been doing, one would type `\`text\`` (see [here](http://meta.stackoverflow.com/editing-help#code)).2012-01-18
  • 0
    It is my pleasant duty to thank you for the third time today, dear @Zev: I actually wanted to know how I could obtain such `elegant grey` ! And I am happy to inform you that I was made aware of your last comments through my newly discovered comment-inbox...2012-01-18

1 Answers 1

5

Let $p\in Z(\mathfrak a) $. We know that $f^r(p)=0$ which implies $f(p)=0$ so $f$ annihilates everything in $Z(\mathfrak a)$.

  • 0
    I'm obviously being very slow: why does $f^r(p) = 0$ imply $f(p) = 0$?2012-01-18
  • 1
    @Matt: Because $f(p) \in K$ and $K$ is a field, so a fortiori an integral domain, and thus $f(p)^r = 0$ implies $f(p) = 0$.2012-01-18
  • 0
    @ZhenLin Does the $f^r$ in Hilbert's Nullstellensatz refer to composition or multiplication?2012-01-18
  • 0
    @Matt, $f^r(p) = (f(p))^r$2012-01-18
  • 0
    @Matt: How could it possibly be composition? It's not a function $K \to K$.2012-01-18
  • 1
    @ZhenLin, composition does make sense for polynomials of one variable but not for several variables, though "multiple composition" $f(g_1,\dots,g_n)$ does (for lack of a better term).2012-01-18