1
$\begingroup$

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 $R=\mathbb{Z}_2$.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – NietzscheanAI Nov 17 '15 at 13:44
  • $\begingroup$ 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 $y$. $\endgroup$ – Ben Grossmann Nov 17 '15 at 13:53
  • $\begingroup$ 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 $x^3 y^4$? $\endgroup$ – NietzscheanAI Nov 17 '15 at 14:00
  • $\begingroup$ 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 $y^4x^3$. $\endgroup$ – Ben Grossmann Nov 17 '15 at 14:02
  • $\begingroup$ 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. $\endgroup$ – Ben Grossmann Nov 17 '15 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.