# Tensor product of two polynomials

Given two polynomials $a$ and $b$ over some ring $R$, what is the explicit definition of their tensor product?

If it's easier to be more concrete, take $R=\mathbb{Z}_2$.

Suppose that $p(x) = \sum a_i x^i$ and $q(x) = \sum b_j y^i$. Then we have $$p(x) \otimes q(y) = \sum_{i,j} a_i b_j (x^i \otimes y^j)$$ The tensor product of the modules $R[x] \otimes R[y]$ is the space of all polynomials of the form $$\sum_{i,j} c_{ij} (x^i \otimes y^j)$$ It may be helpful to think of this as the space of polynomials of the form $$\sum_{i,j} c_{ij} (x^i y^j)$$ where $x$ and $y$ are non-commuting variables.
• Omnomnomnom thanks. A couple of questions for my further clarification, if you'd be so kind: 1. I presume the sum ranges over the Cartesian product of (values of) $i$ and $j$? 2. What is the definition of the $\otimes$ on the RHS? – NietzscheanAI Nov 17 '15 at 13:44
• 1: yes. I guess you could say $i = 0,\dots,m$ and $j = 0, \dots, n$ or something of the like. 2: $v \otimes w$ is the tensor product of the vectors $v$ and $w$. If $v$ and $w$ are "just two vectors", then $v \otimes w$ is "just their tensor product", which is a vector in a larger module. If you want to think of this tensor product as a space of polynomials on non-commuting variables, then $\otimes$ is the product that we use for the symbols $x$ and $y$. – Ben Grossmann Nov 17 '15 at 13:53
• 1. What does non-commuting mean here? `Not necessarily commuting'? 2. Forgive me for laboring this: so for polynomials, $x^3\otimes y^4$ is just $x^3 y^4$? – NietzscheanAI Nov 17 '15 at 14:00
• Yes. Non commuting means not commuting. So, because $x$ and $y$ are non-commuting variables, $xy \neq yx$. Note, then, that elements like $yx$ and $xyx^2$ are not elements of the space of polynomials that we are considering. And yes, you can think of $x^3 \otimes y^4$ as $x^3y^4$, as long as you don't switch these to get $y^4x^3$. – Ben Grossmann Nov 17 '15 at 14:02
• So, quick warning: I assumed that you were thinking of $R[x] \otimes R[y]$ as a module as opposed to an algebra. If you want to multiply two of these elements of these, you need this to be the tensor product of algebras. The multiplication is a bit wonky. In particular, we have $$(x^{n_1} \otimes y^{m_1})(x^{n_2} \otimes y^{m_2}) = x^{n_1 + n_2} \otimes y^{m_1 + m_2}$$ which may seem weird, especially since I told you to think of $x$ and $y$ as non-commuting variables. – Ben Grossmann Nov 17 '15 at 14:19