Product of Polynomials in Several Variables?

Question

Let $p$ and $q$ be the polynomials $\mathbb R$ given by: $$p(x)=\sum_{j=0}^m a_j x^j\quad \textrm{and}\quad q(x)=\sum_{j=0}^n b_j x^j.$$ We know that $$p(x)\cdot q(x)=\sum_{j=0}^{m+n} \left(\sum_{k+\ell=j} a_k b_\ell\right) x^j. $$ What would be the analogous of this formula for polynomails on $\mathbb R^n$?

Obs: Using multi-index notation we might write polynomials $p$ and $q$ on $\mathbb R^n$ in the form: $$p(x):=\sum_{|\alpha|\leq m} a_\alpha x^\alpha\quad \textrm{and}\quad q(x)=\sum_{|\alpha|\leq n} b_\alpha x^\alpha,$$ where $$\displaystyle x^\alpha:=\prod_{j=1}^n x_j^{\alpha_j}.$$ Of course, above I'm supposing $x=(x_1, \ldots, x_n)\in\mathbb R^n$, $\alpha=(\alpha_1, \ldots, \alpha_n)\in\mathbb N_0^n$ and $\mathbb N_0:=\mathbb N\cup\{0\}$.

Conjecture: I think the expression for $p\cdot q$ would be something like: $$p(x)\cdot q(x)=\sum_{|\gamma|\leq m+n} \left(\sum_{\alpha+\beta=\gamma} a_\alpha b_\beta\right)x^\gamma,$$ is it correct?

Answer 1

Your conjecture is correct. Only you use $n$ twice for different things but that's a detail.

This also works for (formal) power series, where it is in fact simpler:

In one variable $$\left( \sum_{i \in \mathbb{N}} a_ix^i \right) \left(\sum_{j \in \mathbb{N}} b_jx^j\right) = \sum_{k \in \mathbb{N}} \left(\sum_{i+j=k}a_ib_j \right)x^{i+j}$$ and in several $$\left(\sum_{\alpha \in \mathbb{N}^s} a_{\alpha}x^{\alpha}\right) \left(\sum_{\beta \in \mathbb{N}^s} b_{\beta}x^{\beta} \right)= \sum_{\gamma \in \mathbb{N}^s} \left(\sum_{\alpha +\beta = \gamma}a_{\alpha}b_{\beta} \right)x^{\alpha+\beta}.$$

Note that for this to work for powerseries without issue it is important that $\alpha +\beta = \gamma$ has only finitely many solution for fixed $\gamma$. Indeed, one could do the same for any commutative semigroup where one has that property, but I guess I digress now.

You can recover the polynomials as a the special where only finitely many of the coeeficients are non-zero.