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Let $R$ be a commutative ring with $1$ and let $A$ be a commutative $R$-algebra. We view $A\otimes A$ also as $R$-algebra, the multiplication on generatrs being given by $(a\otimes b)(a'\otimes b')= aa'\otimes bb'$.

Denote by $m: A\otimes A\rightarrow A$ the product homomorphism, i.e. the map that maps a generator $a\otimes b$ to $ab$.

My question is: why is the kernel of $m$ generated by the elements $a\otimes 1 - 1\otimes a$ where $a$ runs through $A$?

The answer seems to be so easy that I could not find it anywhere in books. Alas, it is not clear to me :(

Thank you for your help!

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Is $R$ implicitly assumed to be an integral domain? Otherwise, if $ab=0$, how would you write $a\otimes b$ in terms of these generators? – Espen Nielsen Dec 19 '12 at 16:45
@espen180: You mean if $A$ has no zero divisors? No, neither $R$ nor $A$ are implicitly assumed to have no zero divisors. Note that the kernel of $m$ is an ideal. For $ab=0=ba$, we could write $a\otimes b = (a\otimes 1-1\otimes a)(1\otimes b - b\otimes 1) - (b\otimes 1-1\otimes b)(1\otimes a)$, so it lies in the ideal generated by the $a\otimes 1-1\otimes a$, hence in the kernel. – Sh4pe Dec 19 '12 at 17:05
@espen180: even shorter: $a\otimes b=(a\otimes 1-1\otimes a)(1\otimes b)$... – Sh4pe Dec 19 '12 at 17:12
Of course, my mistake. But doesn't this prove the theorem, then? Since it is obvious that the above generated set is included in the kernel, and you just showed the reverse inclusion? – Espen Nielsen Dec 19 '12 at 17:30
This was asked and answered at… – Mariano Suárez-Alvarez Dec 19 '12 at 17:35
up vote 1 down vote accepted

Hint: Show that $a\otimes b - ab\otimes 1$ is in $I$, the ideal generated by elements of the form $x\otimes 1 - 1\otimes x$. This can easily be done by noting that $a\otimes b = (a\otimes 1)(1\otimes b)$, then substituting $1\otimes b = b\otimes 1 - (b\otimes 1 - 1\otimes b)$.

From here, show that every element of $A\otimes A/I$ can be written as $x\otimes 1 + I$.

Since $I$ is a sub-ideal of the kernel of your morphism, your morphism factors:

$$A\otimes A \to A\otimes A/I \to A$$

If $I$ is not the kernel, then $A\otimes A/I \to A$ is not $1-1$. But every element of $A\otimes A/I$ can be written as $x\otimes 1 + I$ and it must be sent to $x$. So this is necessarily $1-1$.

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Easy: for $a\otimes b$ in the kernel, write $a\otimes b = (a\otimes 1-1\otimes a)(1\otimes b)$ and be done


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There are elements in the kernel which are not elementary tensors, though. To get an example, pick pretty much any integral domain $A$, like $k[X]$: in this last example you can easily check that there is no elementary tensor in the kernel of the multiplication map! – Mariano Suárez-Alvarez Dec 19 '12 at 17:36
The elements of $A\otimes A$ are not only of the form $a\otimes b$ – Thomas Andrews Dec 19 '12 at 17:44
(That is pretty much my first sentence!) – Mariano Suárez-Alvarez Dec 19 '12 at 17:45
Yeah, didn't update the page. – Thomas Andrews Dec 19 '12 at 17:53

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