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Let $k$ be an algebraically closed field and $A$ be a $k$-algebra (commutative). Let $\phi$ be the map

$$ \phi : A \otimes_{k} A \to A $$

given by the algebra multiplication.

I would like to know the kernel of this map.

One sees that the elements of the form $a \otimes 1 - 1 \otimes a$ lie in the kernel. Is the ideal $I$ generated by them the full kernel?

This arose from when I was studying linear algebraic groups and trying to prove that affine varities are varities (separation axiom holds). It is claimed in Springer's book that $\Delta_{X}$ is the set of zeros of the ideal $I$, which is obviously true (Nullstellensatz), but he also claims that $I$ is the kernel of $\phi$ which I do not see. Of course, for him $A $ is a quotient of polynomial algebra; so if that helps to prove, it might be assumed.

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Let $z = \sum\limits_i a_i \otimes b_i$ be in the kernel, so

$$\sum\limits a_ib_i = 0$$

Then $$ \sum\limits_i (a_i \otimes 1)(b_i \otimes 1 - 1 \otimes b_i) $$ $$= \sum\limits_i [ a_ib_i \otimes 1 - a_i \otimes b_i] = \left(\sum\limits_i a_ib_i\right) \otimes 1 - \sum\limits_i (a_i \otimes b_i) = - z $$