# Total degree of a product of polynomials

Note: This is not a duplicate of this question. The accepted answer there gives some answer to my question here, but the question is not a duplicate...

I'm currently brushing up on some algebra with "Abstract Algebra" by Grillet, Second Edition.

We consider, for a unital Ring $R$ and a family of sets $(X_i)_{i\in I}$ the polynomial ring $$R[(X_i)_{i\in I}]$$

I don't think the details of this construction are too important here; suffice it to say that we are dealing with the free $R$-algebra over the free commutative monoid of all monomials $$X^k=\Pi_{i\in I}\, X^{k_i}\,,\; k \in \mathbb{N}^I\,, k_i = 0\, \text{ for almost all } i$$
with multiplication $X^k X^l = X^{k+l}$.

The degree of a polynomial $0 \neq A \in R[(X_i)_{i\in I}]$ with coefficients $a_k$ is defined as $$\max\left\{\sum_{i \in I} k_i;\, k\in I^\mathbb{N}\,,\, k_i = 0\, \text{ for almost all } i\, \land \, a_k \neq 0\right\}$$

I generally put quite a lot of trust into textbooks as carefully written as this one, but the following statement on page 127 seems wrong to me and I am stuck trying to prove it:

Proposition 6.4. For all $A, B \neq 0$ in $R[(X_i)_{i\in I}]$:
...
(4) if $R$ has no zero divisors, then $\deg(\,AB\,) = \deg A + \deg B$

What I have proven is $\deg(\,AB\,) \leq \deg A + \deg B$, which was part (3) of the proposition.

However, part (4) baffles me. I started out expecting this to be easy but got stuck; and now I'm really a little too exhausted because I couldn't let go and kept on thinking about how to prove this.
I'm actually wondering if this is actually wrong and was just a copy-paste error from the corresponding page 121 for the univariate case, where we have the exact same statement, safe for there only being one variable.
A counterexample is probably not that hard, but I'm really too tired right now, and then I actually already got sidetracked just brushing up on basics required for my bachelor's thesis with this already, in the first place.

Why I feel like this isn't right: For $k \in I^\mathbb{N}$ with finite support, the $k$-th coefficient of $AB$ is

$$(AB)_k = \sum_{(m,\,n) \in (I^\mathbb{N})^2;\,m + n = k} a_m b_n$$

where the $a_i, b_i$ are the coefficients of $A$, respectively $B$. In the univariate case, the degree formula holds if $R$ has no zero divisors, because the leading coefficient of the product is the product of the leading coefficients.
However, in the multivariate case, each coefficient can be a sum of products:

Picking indices $\nu\,,\, \mu$ with $\sum_i \nu_i = \deg A$ and $\sum_i \mu_i = \deg B$, I don't see how $R$ having no zero divisors would lead to the coefficient $$(AB)_{\nu + \mu} = \sum_{(m,\,n) \in (I^\mathbb{N})^2;\,m + n = \nu + \mu} a_m b_n$$ being nonzero, although it seems obvious that $AB \neq 0$ if we look at the $i$-degree $$\max\left\{k_i;\, k\in I^\mathbb{N}\,,\, k_j = 0\, \text{ for almost all } j\, \land \, a_k \neq 0\right\}$$ for some $i \in I$ (should I be wrong with this reasoning, please tell me), so that $R[(X_i)_{i\in I}]$ is a domain, if $R$ is.

So, am I right in assuming this part of the proposition quoted above is wrong?

I'm all out of creative mojo, so if someone could just plug me an easy counterexample, I'd be very, very happy... if the statement is actually true, I'd appreciate some help in proving it.

• Jan 14 '18 at 19:09
• I voted to reopen this question. As the OP writes, it is related, but not a duplicate. Also, the answer given there is of some use, but it is not a complete answer to this question. This should not have been closed. Jun 18 '20 at 15:48

As all polynomials have finitely many terms, we may as well assume that we are working in $R[X_1,\dots,X_n]$. Order the monomials $X_1^{a_1}X_2^{a_2}\cdots X_n^{a_n}$ in the graded lexicographic ordering. In other words one monomial is larger than another if the first has higher total degree or if they have equal total degree and the first is larger in the lexicographic ordering. The point is that this order, $X_1^{a_1}\cdots X_n^{a_n}>X_1^{b_1}\cdots X_n^{b_n}$ and $X_1^{c_1}\cdots X_n^{c_n}>X_1^{d_1}\cdots X_n^{d_n}$ implies $X_1^{a_1+c_1}\cdots X_n^{a_n+c_n}>X_1^{b_1+d_1}\cdots X_n^{b_n+d_n}$. Define the leading coefficient of $f$ as the non-zero coefficient attached to the largest monomial in $f$. Then the leading coefficient of $fg$ is that of $f$ times that of $g$, and is nonzero when $R$ has no zero divisors. Moreover it is of degree equal to that of the sum of the degrees of $f$ and $g$.
• Ah, great way to prove this :) Didn't expect to have to introduce an ordering. I like the details... essentially, one would pick all the $i_1\,,\,\dots\,,\,i_l \in I$ for which $f$ and $g$ have nonzero coefficients and define a lexicographical ordering for those (which is implied by the 'we may as well assume'-statement). So, the proof uses infinetely many orderings, implicitly :) Jan 18 '18 at 20:49
• .. or rather: 'all the $i_1, \, \dots\,, i_l \in I$ for which the respective $X_j$ appear in monomials that have nonzero coefficients', to be more precise... Jan 18 '18 at 20:56