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Let $X$ be an affine variety over a field $k$ of characteristic $p$. Let $A$ be the set of morphisms $X\to \mathbb A_1$. Let $B\subset A$ be a sub-algebra. Suppose $f\in A$ is such that

$$ \bigcap_{g\in B} \ker g \subset \ker f.$$

Then if $A/B$ has no $p$-torsion, we have that $f\in B$. My proof goes as follow: For any $h\in A$, let $\bar h$ be the image of $h$ in $A/B$. Then,

$$ X = \bigcap_{g\in B} \ker\bar{g} \subset \ker\bar{f}.$$

Thus $X = \ker\bar f$, which implies that $\bar f = 0$, so $f\in B$. I did not use $p$-torsion. Can you please tell me what is wrong with this proof and why $p$-torsion is so important?

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I changed $\displaystyle\cap_{g\in B}\ker g$ to $\displaystyle\bigcap_{g\in B}\ker g$. When \bigcap rather than \cap is used in a "displayed" (as opposed to "inline") setting, it is not only bigger, but the subscript appears directly under it. I think \cap is intended for situations like $A\cap B$ and $A\cap\cdots\cap B$. – Michael Hardy Feb 17 '13 at 17:51
What does kernel mean? – user27126 Feb 17 '13 at 17:58
Hi Sanchez, I change the question a little bit. Does it make more sense? – user61408 Feb 17 '13 at 18:03
As far as I can tell, there is no good way to define $\ker\bar{f}$. How are you defining it...? – Nehsb Feb 17 '13 at 18:35
@user61408, I'm still confused. I guess that when you say kernel, you really mean the zero set of the function, but then kernel of something in $A/B$ doesn't even make sense. – user27126 Feb 17 '13 at 19:48

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