Understanding of an example of "extending scalars" The following is an example in the Abstract Algebra by Dummit and Foote:


I don't understand in this example why $\iota$ is an isomorphism. By Theorem 8, I can get
$$
id_N=\Phi\circ\iota
$$
which implies that $\iota$ is injective and $\Phi$ is surjective. Why is $\iota$ surjective or $\Phi$ is injective?
 A: Sorry, my previous answer was incorrect: I attempted to be too fancy.
It looks like a straightforward approach is probably best. Take a simple tensor in $R \otimes_R N$ which will be of the form $r \otimes n = r(1 \otimes n) = r \iota (n)$. Arbitrary elements are finite sums of simple tensors so since $\iota$ is a homomorphism, we are done. 
A: Let $\varphi: R \times N \to N$, $(r,n) \mapsto rn$ be the $R$-bilinear multiplication map, and let $\Phi: R \otimes_R N \to N$ be the map induced by the universal property of the tensor product, so $rn = \varphi(r,n) = \Phi(r \otimes n)$.  We show that the map
\begin{align*}
\iota: N &\to R \otimes_R N\\
n &\mapsto 1 \otimes n
\end{align*}
and $\Phi$ are inverse, hence are isomorphisms.  Note that
$$
\Phi(\iota(n)) = \Phi(1 \otimes n) = \varphi(1,n) = 1 \cdot n = n
$$
and since $\Phi$ and $\iota$ are $R$-linear, then
\begin{align*}
\iota\left(\Phi\left(\sum_\alpha r_\alpha \otimes n_\alpha \right)\right) &= \iota\left(\sum_\alpha \Phi(r_\alpha \otimes n_\alpha)\right) = \iota\left(\sum_\alpha r_\alpha n_\alpha \right) = \sum_\alpha \iota(r_\alpha n_\alpha)\\
&= \sum_\alpha 1 \otimes r_\alpha n_\alpha = \sum_\alpha r_\alpha \otimes n_\alpha \, .
\end{align*}
(Every tensor is a finite sum of simple tensors, so here the sum is over $\alpha$ in some finite index set $I$.)  Thus $\Phi$ and $\iota$ are (mutually inverse) isomorphisms.
