I am to use the following corollary to show $R[x]\otimes R[y]\cong R[x,y]$
Corollary 11, Chapter 10 Section 3, Dummit and Foote
I have come up with the map $\iota^{'}:R[x]\times R[y]\rightarrow R[x,y], \iota^{`}((a,b))=ab$. I have shown that this map is R-balanced and I have shown that the image generates $R[x,y]$ ($R$ has unity). I am confused about the second part of the corollary. First, I do not understand the reference back to Theorem 10, because in Theorem 10 we had three maps, and here we do not. Theorem 10 is below.
Theorem 10, Chapter 10 Section 3, Dummit and Foote
where $\iota:R[x]\times R[y]\rightarrow R[x]\otimes R[y].\iota((a,b))=a\otimes b$
The second thing that I am confused about is, how do I show that all homomorphisms satisfy this property?
Here's a sketch of the idea: Suppose $\varphi:R[x] \times R[y] \longrightarrow L$ is $R$-balanced. We need to prove the existence of a unique group homomorphism $\tilde{\varphi}:R[x,y] \longrightarrow L$ such that $\tilde{\varphi} \circ \iota' = \varphi$. What could this map $\tilde{\varphi}$ possibly be? The answer is clear for monomials: $\tilde{\varphi}(rx^jy^k) = \varphi(rx^j,y^k)$, which is equal by $R$-balancedness to $\varphi(x^j,ry^k)$. Since every element of $R[x,y]$ can be expressed as a finite $R$-linear combination of monomials. We thus define: $\tilde{\varphi}(f(x,y)) = \tilde{\varphi}(\sum_{j,k} r_{j,k}x^jy^k) = \sum_{j,k} \varphi(r_{j,k}x^j,y^k)$. You'll have to verify that this defines a homomorphism and that it is unique (i.e., if some other map $\psi:R[x,y] \longrightarrow L$ satisfies $\psi \circ \iota' = \varphi$, then $\psi = \tilde{\varphi}$), but that shouldn't be too hard.
By the way, this approach is very common in algebra. Rather than constructing an explicit isomorphism between two objects (in this case, between $R[x,y]$ and $R[x] \otimes_R R[y]$) we instead show that the one object possesses a property that universally characterizes the other object (here, it is the unique factorization property.) Since every $R$-balanced map $R[x] \times R[y] \longrightarrow L$ uniquely factors through $\iota':R[x] \times R[y] \longrightarrow R[x,y]$, and since the only abelian group with this property (up to isomorphism) is $R[x] \otimes_R R[y]$, we can conclude that the two are isomorphic. Pretty slick! :)