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Let $X$ be an affine variety and $f,g\in \mathcal{O}(X)$ such that $\lbrace f=0\rbrace = \lbrace g= 0\rbrace$. How does it follow by the Nullstellensatz that $f^n = g^m$ for some $m,n\in \mathbb{N}$? This is claimed in a proof, but I don't see the reason.

If it is not true in general, are there additional conditions under which the statement is true?

Many thanks in advance!

1 Answers 1

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It doesn't follow because it's not true. For example, let $f = x^2 y, g = x y^2$. (This shows that even the weaker claim $(f^n) = (g^m)$ for some $n,m$ is false. The strong claim is false for silly reasons like the coefficients not matching up; for example, let $f = x, g = 2x$ and let the underlying field have characteristic zero.)

The correct statement is that the Nullstellensatz implies that $(f), (g)$ have the same radical. In particular, $f^n \in (g)$ and $g^m \in (f)$ for some $n, m$.