Tensor Products - How two polynomial rings in one variable give one polynomial ring in two variables! (huh?)

Let $R$ be a ring. The set of all polynomials in the variable $x$ with coefficients in the ring $R$ is denoted by $R[x]$. An example of an element in the set $R[x]$ would be $x^3 + 5$.

Some of you would know that $R[x]$ is a module over $R$!

Now, let’s look at the polynomial rings $R[x]$ and $R[y]$ - they’re essentially the same except one consists of polynomials in the variable $x$, and the other in $y$. How are the two related to the polynomial ring $R[x,y]$?

To make things easier, let’s fix $R$ to be our favourite ring $\mathbb{Z}$. If I pick an element in $R[x]$, say $x^3 + 5$, and an element from $R[y]$, say $2y$, and multiply the two, then I get the polynomial $(x^3 + 5)(2y) = 2x^3y + 10y$, which is polynomial in $x$ AND $y$, i.e. an element in $R[x,y]$.

OK - so multiplying an element in $R[x]$ with an element in $R[y]$ seems to give us an element in $R[x,y]$. So maybe we can define a multiplication map: $$\phi\colon R[x]\times R[y]\rightarrow R[x,y],$$ which sends $(f(x),g(y))$ to $f(x)\cdot g(y)$.

Can we dare to hope that $\phi$ is a bijection (or maybe an isomorphism of $R$-modules)?

Well, if you’ve had any experience in solving differential equations, and therein a technique called \textit{separation of variables}, you’d know this map wouldn’t be surjective. Indeed, not every polynomial in two variables $x$ and $y$ can be split into two polynomials of one variable each - one in $x$ and the other in $y$. For instance, $x^2y^2 + y$ can’t be split into the product of a polynomial in $x$ and one in $y$.

Is it at least injective? Well, the product of $x$ and $2y$ is the same as the product of $2x$ and $y$, meaning that the pairs $(x,2y)$ and $(2x,y)$ map to the same element under $\phi$, meaning $\phi$ isn’t even injective.

So $\phi$ is neither injective nor surjective - very disappointing for a map so promising. Is there ANYTHING nice about $\phi$?

Let’s look at what stopped $\phi$ from being a bijection.

\begin{enumerate} \item $\phi$ is not \textbf{surjective} because $R[x,y]$ has polynomials other than those that are a product of a polynomial in $x$ and one in $y$. However, these nice polynomials in the image of $\phi$ that look like $f(x)g(y)$ GENERATE all the polynomials in $R[x,y]$ i.e. any polynomial in two variables can be written as a finite sum of polynomials of the form $f(x)\cdot g(y)$. For instance, in the example from above, $x^2y^2 + y = (x^2)(y^2) + (1)(y)$. \item $\phi$ is not \textbf{injective} because $\phi(2x,y) = \phi(x,2y)$. In fact, we know exactly when two pairs of polynomials will map to the same thing under $\phi$: for any $c\in R$, \begin{equation}\label{scalar} \phi(cf(x),g(y)) = \phi(f(x),cg(y)) \end{equation} . \end{enumerate}

In addition, thanks to the distributive property of polynomials multiplication, we have $\phi(f_1(x)+f_2(x),g(y)) = \phi(f_1(x),g(y)) + \phi(f_2(x),g(y))$, and a similar “linearity” in the second component. This, along with the condition described above in Equation \ref{scalar} make the map $\phi$ into what’s called an $R$- \textit{bilinear} map. These properties are what make $\phi$ VERY special as we shall see now!

Suppose we had another bilinear map $F\colon R[x]\times R[y]\rightarrow A$, where $A$ is any $R$-module, or even just an abelian group. Let us try to define a group homomorphism $\overline{F}\colon R[x,y]\rightarrow A$ using $F$.

As observed above, $R[x,y]$ is generated by polynomials of the form $f(x)g(y)$, and so it is sufficient to describe where these generators are mapped to under $\overline{F}$ to completely determine the map. We can therefore define $\overline{F}(f(x)g(y)) = F(f(x),g(y))$. The product $f(x)g(y)$ is of course equal to $\phi(f(x),g(y))$, and therefore we have $\overline{F}(\phi(f(x),g(y))) = F(f(x),g(y))$.

The properties of $\phi$ noted above make sure that this map is well-defined i.e. even though the definition of $\overline{F}$ depends on how you split a polynomial - say $2xy$ - as a product, it won’t send the same element in $R[x,y]$ to two different elements in $A$ based on how you split it. For instance, if you write $2xy$ as $2x\cdot y$, then $\overline{F}$ would send it to $F(2x,y)$, and if you split it as $x\cdot 2y$, then $\overline{F}$ would send it to the element $F(x,2y)$ in $A$. But since $\phi$ and $F$ are bilinear, these two elements in $A$ are equal, so there’s no confusion (check this)!

This gives rise to the following commutative diagram (i.e. if you pick an element in $R[x]\times R[y]$, it’ll land on the same element in $A$ no matter which path you take) -

$$\begin{tikzcd} R[x]\times R[y]\arrow[d,"\phi"]\arrow[r,"F"] & A\\ R[x,y]\arrow[ur,"\overline{F}"] \end{tikzcd}$$

Since this holds for \textit{any} choice of abelian group $A$ with a map $F\colon R[x]\times R[y]\rightarrow A$, the above property of $R[x,y]$ is called a \textit{universal property}. This group $R[x,y]$ becomes special because it turns bilinear maps from $R[x]\times R[y]$ into group homomorphisms from $R[x,y]$. We call $R[x,y]$ the \textbf{tensor product} of $R[x]$ and $R[y]$ and denote it by $R[x]\otimes_R R[y]$ (the subscript $R$ refers to the fact that it is elements $c\in R$ that can be moved around between the two components in the map $\phi$ as in Equation \ref{scalar}) .

We are now (hopefully) motivated enough to define the tensor product of two objects- say modules over a commutative ring $R$ -

Let $M$ and $N$ be two modules over $R$. The tensor product $M\otimes_R N$ of $M$ and $N$ over $R$ is an abelian group with an $R$-bilinear map $\phi\colon M\times N\rightarrow M\otimes_R N$ which satisfies the following universal property - for any abelian group $A$ and an $R$-bilinear map $F\colon M\times N\rightarrow A$, there exists a unique group homomorphism $\overline{F}\colon M\otimes_R N\rightarrow R$ such that the following diagram commutes:

$$\begin{tikzcd} M\times N\arrow[d,"\phi"]\arrow[r,"F"] & A\\ M\otimes_R N\arrow[ur,"\overline{F}"] \end{tikzcd}$$

\textit{\textbf{Bonus Question:}} The whole discussion started with the observation that the product of a polynomial $f(x)$ in $x$ with a polynomial $g(y)$ in $y$ gives a polynomial in two variables $x$ and $y$ i.e. an element in the set $R[x,y]$. But we could have also taken the SUM of $f(x)$ and $g(y)$ and still ended up with an element in $R[x,y]$. Could this have been used to get to the tensor product? Or does the sum map give us some other universal property?

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