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We assume that $\Bbb k$ is an algebraically closed field.

Let $X \subset \Bbb A^n$ be an affine $\Bbb k$-variety , let's consider $ \mathfrak A_X \subset \Bbb k[t_1,...t_n]$ as the ideal of polynomials that vanish on $X$. Given a closed subset $Y\subset X$ we associate the ideal $ a_Y \subset k[X]$ defined by $ a_Y = \left\{ {f \in k[X];f = 0\,on\,Y} \right\} $. I'm reading " Basic Algebraic Geometry of Shafarevich" and it says that " It follows from Nullstellensatz that $Y$ is the empty set if and only if $a_Y = K[X] $ But I don't know how to prove this . Maybe it's trivial , but I need help anyway.

After knowing if the result is true here, I want to know if it's true in the case of quasiprojective varieties, but first the affine case =)

It remains to prove that if $ Y\subset X $ then $$ Y = \phi \Rightarrow a_Y = k\left[ X \right] $$ For that side we need Hilbert Nullstelensatz, but I don't know how to use it.

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I have replaced your $U_X$ by $\mathfrak A_X$, the letter that Shafarevich actually uses: it is a Gothic A ( Fraktur A in German).I have also corrected the spelling of Nullstellensatz. –  Georges Elencwajg Jan 4 '13 at 23:31
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I'm not sure why the Nullstellensatz is necessary here. If every function vanishes on a set $Y$ (including all constant functions) then the set $Y$ must obviously be empty; else the function $f(x) = 1$ would be non-vanishing at some point $x$. Conversely if the set $Y$ is empty then every function attains 0 at every point of $Y$. –  user54535 Jan 4 '13 at 23:35
    
Yes... I'm not sure why they need Nullstelensatz :/! –  Daniel Jan 4 '13 at 23:38
    
You are absolutely right @User 24601, the Nullstellensatz is completely irrelevant here: +1. It is reassuring for us mere mortals that even Shafarevich could make such an egregious mistake! –  Georges Elencwajg Jan 4 '13 at 23:41
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Dear Daniel, you call your base field $k, \mathbb k$ and $K$. Please, make up your mind! –  Georges Elencwajg Jan 4 '13 at 23:46

1 Answer 1

The Nullstellensatz is unnecessary here. If every function vanishes on a set Y (including all constant functions) then the set Y must obviously be empty; else the function f(x)=1 would be non-vanishing at some point x. Conversely if the set Y is empty then every function attains 0 at every point of Y.

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We want to prove the following: $$ Y = \phi \Leftrightarrow a_Y = k\left[ X \right] $$ We proved one side of the equality , more specifically $ \left( \Leftarrow \right) $ proving the contrapositive. For the other side we probably need Nullstelensatz, in fact this is more general than Nullstelensatz, to see this only consider $ Y\subset X $ but $X=\Bbb A^n$ , then $ k[X]=k[t_1,...,t_n]=k[T] $ , and in this case $ a_Y = I(Y) $ (the polynomials that vanish on Y ) thus $ Y = \phi \Leftrightarrow a_Y = I\left( Y \right) = k\left[ X \right] = k\left[ T \right] $ which is precisely hilbert –  Daniel Jan 8 '13 at 12:37
    
Daniel, this is not more general than the Nullstellensatz. And both implications have been proven in the answer above. If $I(Y) = k[X]$, then every function vanishes on $Y$ -- including the function that attains the value $1$ at every point, for instance. (This corresponds to the unit element of $k[X]$.) Now, what does this say about $Y$? Remember that many logical statements are vacuously true for the empty set. –  user54535 Jan 8 '13 at 15:22

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