I am (re-)reading the section on tensor products of $R$-algebras in Atiyah-Macdonald, where $R$ is a commutative ring, and I am not sure about the given definition of the $R$-module structure on the tensor product.
Let $R$ be a commutative ring, and let $(S,f)$ and $(T,g)$ be $R$-algebras. Then, it is claimed that the tensor product $V = S \otimes_R T$ has an $R$-algebra structure due to the homomorphism:
$$R \to V, \;\; r \mapsto f(r)\otimes g(r)$$
However, I don't see how the canonical map $\psi :S \times T \to S \otimes T$ can be bilinear if this is right.
Let $r \in R, (s,t) \in S \times T$. Then, on the one hand:
$$\psi(r*s,t) = \psi(f(r)s,t)=f(r)s\otimes t$$
But on the other:
$$r * \psi(s,t) = r * (s \otimes t) = (f(r) \otimes g(r)) (s \otimes t) = (f(r)s \otimes g(r) t)$$
Therefore, if $\psi$ is bilinear, we must have that:
$$f(r)s\otimes t \stackrel{?}{=} (f(r)s \otimes g(r) t)$$
which does not seem correct to me.
Have I misunderstood something, or is there an error here? If there is, please could I know the true $R$-module structure on $V$?
As mentioned by Mariano in the comments, this is an error in Atiyah-MacDonald, which you can verify because this map is not linear, and the correct map is $r \mapsto f(r) \otimes 1 = 1 \otimes g(r)$.
This confusion could have been avoided by taking a different definition of an $R$-algebra. Instead of defining it as a ring $A$ equipped with a unit map $R \to Z(A)$, which I've always thought of as sort of inelegant, you can define it to be an $R$-module equipped with an $R$-bilinear multiplication making it a ring. The unit map comes from multiplying the unit by scalars in $R$. If $A, B$ are two $R$-algebras then the tensor product $A \otimes_R B$ inherits an $R$-algebra structure simply because it is a tensor product of two $R$-modules so inherits an $R$-module structure, and the natural multiplication on $A \otimes_R B$ is $R$-bilinear.
Since the unit of $A \otimes_R B$ is the tensor product $1 \otimes 1$, we recover the unit map by multiplying the unit by scalars in $R$, and by bilinearity this gives $r(1 \otimes 1) = r \otimes 1 = 1 \otimes r$ as expected.
Abstractly the significance of this definition is that it's a special case of the much more general notion of a monoid object in a monoidal category, in this case the monoidal category of $R$-modules with respect to tensor product. This general definition explains, among other things, why we don't have a (straightforward) notion of $R$-algebra if $R$ is noncommutative: it's because if $R$ is noncommutative the category of (left or right) $R$-modules no longer possesses a tensor product.