Natural homomorphism from $A$ and $B$ to $A \otimes B$ 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.
 A: 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.
A: 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. 
