Suppose $A$ and $B$ are two $R-$algebras, where $R$ is a commutative ring with $1$. There are natural algebra homomorphisms $\ a \to a \otimes 1 \ $ and $ \ b \to 1 \otimes b \ $ from $A$ and $B$ to $A \otimes_{R} B$.

Are these homomorphisms injective?

Consider the first homomorphism. If $a_1 \neq a_2$ then we want $(a_1 - a_2) \otimes 1 \neq 0$. If there is an $R-$bilinear map $f: A \times B \to M$, where $M$ is some $R-$module, such that $f(a_1 - a_2, 1) \neq 0$ then we are done. But how can one find such $f$?

UPDATE: As it was pointed out, it is not true in general. But when is it true?

It seems that if $A$ and $B$ are real or complex vector spaces, both homomorphisms are injective. Furthermore, when $A$ is a $\mathbb{Z}$-tosrion-free ring and $B$ is a field of characteristic $0$, then $a \to a \otimes 1$ seems to be injective.

  • $\begingroup$ Note that the definition of tensor product requires a choice of base ring. If the base ring is a field, e.g. $\mathbb R$ or $\mathbb C$, then the canonical tensor product maps are injective. $\endgroup$ – Dustan Levenstein Nov 24 '17 at 13:28
  • $\begingroup$ Tsemo's example uses the base ring $\mathbb Z$. $\endgroup$ – Dustan Levenstein Nov 24 '17 at 13:28
  • $\begingroup$ @DustanLevenstein I see no contradictions. I consider the product $A \otimes_{R} B$. In my first example $R = \mathbb{R}$ or $\mathbb{C}$ and in the second one $R = \mathbb{Z}$. $\endgroup$ – user128245 Nov 24 '17 at 17:47
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    $\begingroup$ I don't know what you mean by contradictions. I was just pointing that out because you said "It seems that if $A$ or $B$ are real or complex vector spaces, both homomorphisms are injective", which is only true if you're fixing the base ring to be respectively $\mathbb R$ or $\mathbb C$. For example, $\mathbb C[x]/x^2 \otimes_{\mathbb C[x]} \mathbb C[x]/(x-1)^2 = 0$. $\endgroup$ – Dustan Levenstein Nov 24 '17 at 18:05

We can approach the situation abstractly as follows. These maps are tensor products of inclusions of scalar multiples of the identities $R \to A$ and $R \to B$ into $A$ and $B$, which are then tensored by $B$ and $A$ respectively. So, a sufficient condition for the tensored map

$$R \otimes_R B \cong B \mapsto A \otimes_R B$$

to be injective is first that the original map $R \to A$ is injective and second that $B$ is flat as an $R$-module, and similarly the other way.

Both conditions are automatic if $R$ is a field but not in general. If $R \to A$ is not injective then the identity is a torsion element in $A$ as an $R$-module, so a sufficient condition for this to not happen is that $A$ is torsion-free; also, if $B$ is flat then it is torsion-free. These assumptions are sufficient if $R$ is a Dedekind domain, since then every torsion-free module is flat.


No,$\mathbb{Z}/2\otimes_{\mathbb{Z}}\mathbb{Z}/3=0$, so $\mathbb{Z}/2\rightarrow \mathbb{Z}/2\otimes_{\mathbb{Z}}\mathbb{Z}/3$ is not injective.

  • $\begingroup$ Agreed: this is my automatic example for this matter. $\endgroup$ – Lubin Nov 24 '17 at 20:25

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