Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that we have a commutative, associative ring $R$ which we use to generate the polynomial $R[x]$. Then $p(x) \in R[x]$ is of the form $p(x) = \sum_{i=0}^n a_i x^i$ for some $n \in \mathbb{N}$.

Now, I wish to understand why $R[x]$ is itself a commutative, associative ring. I am happy showing all of the necessary properties to show this to be the case, except for associativity of multiplication.

I understand that given $p(x), q(x) \in R[x]$, such that

$$p(x) = \sum_{i=0}^n a_i x^i, \ \ q(x) = \sum_{i=0}^m b_i x^i$$

for some $m, n \in \mathbb{N}$, where $a_i, b_i \in R, \forall i$, we may (quite naturally) define their product

$$p(x)q(x)= \left(pq \right)(x):=\sum_{k=0}^{n+m} \sum_{i+j=k, \ i,j \in \mathbb{N}_0} a_i b_j x^k.$$

However, I get rather lost when I introduce another polynomial, $r(x) \in R[x]$ such that

$$r(x) = \sum_{i=0}^o c_i x^i$$

for some $o \in \mathbb{N}$, where $c_i \in R, \forall i$ and attempt to show that

$$\left(p(x) q(x) \right) r(x) = p(x) \left(q(x) r(x) \right).$$

Would any kindly soul be able to show me how?

share|cite|improve this question
I think that these proofs are harder to write down than to be thought. Suggestion: write down an explicit example and convince yourself that there are no problems. – Valerio Capraro Oct 31 '11 at 23:58
Why do you get lost? With the approach you've taken, it does take a little perseverance, but that's about it. Write explicitely what $(pq)r$ is. Do the same for $p(qr)$. Compare! – Mariano Suárez-Alvarez Oct 31 '11 at 23:58
I think that Valerio's suggestion is a bad one: one should be able to write complete proofs of such basic facts. "Convincing" oneself should mean "convince oneself that one can actually prove" and, until one has considerable experience (I have not gotten there...) that means in most cases write the complete proof. I have encountered uncountably many instances of people who'd convinced themselves of things they could not prove... – Mariano Suárez-Alvarez Nov 1 '11 at 0:00
Can we consider polynomials as functions $R \to R$ and in this case the result simply follows from the associativity of mulplication in $R$? – user17090 Nov 1 '11 at 0:06
I'm one of them, sure :) – Valerio Capraro Nov 1 '11 at 0:09

Instead of directly showing that two expressions for triple products with different orders are the same, it might be both easier and more illuminating to show that the $k$-th coefficient in the product of any number of polynomials taken in any order is the sum of all products of coefficients, one from each factor, whose powers add up to $k$. This implies associativity.

We can prove this by structural induction over the factors. So let some multiple product of polynomials be given, and focus on the "outermost" multiplication, the one to be applied last. We have

$$p(x) = \sum_{i=0}^n a_i x^i, \ \ q(x) = \sum_{i=0}^m b_i x^i\;,$$

where $p(x)$ and $q(x)$ are the results of carrying out all the other multiplications. By the induction hypothesis, $a_i$ is the sum of all products of coefficients, one from each factor in the left factor, whose powers add up to $i$, and the same for $b_i$ and the right factor. Now the $k$-th coefficient in the product is by definition

$$\sum_{i+j=k, \ i,j \in \mathbb{N}_0} a_i b_j\;.$$

Clearly when we multiply all these products out using distributivity, we'll get every product of coefficients, one from each factor, whose powers add up to $k$, and we'll get it exactly once. That completes the induction.

share|cite|improve this answer

I agree with the structural approaches suggested in the other answers. But I don't see where's the problem with a direct computation? It isn't horrible at all and actually also works for formal power series.

I will write $p_i$ for the coefficient of $x^i$ in $p$. We have by definition

$$(p(qr))_n = \sum_{i+j=n} p_i (qr)_j = \sum_{i+j=n} p_i \sum_{u+v=j} q_u r_v = \sum_{i+j=n} ~\sum_{u+v=j} p_i (q_u r_v)$$

By a simple substitution this equals

$$\sum_{i+u+v=n} p_i (q_u r_v).$$ Now use associativity in $R$ and the same reasoning backwards, then you get $((pq)r)_n$.

share|cite|improve this answer

I've been wondering about the same thing, two years later. Here's a proof outline I came up with that limits the horrible sums at the expense of requiring a lemma and three different inductive steps. In the following, all polynomials are over a ring $R$.

Lemma: Let $p(x)$ and $q(x)$ be polynomials, let $a\in R$ and $n\in \Bbb N_{>0}$. Then $ax^n(p(x)q(x)) = (ax^np(x))q(x)$, $p(x)(ax^nq(x))=(p(x)ax^n)q(x)$, and $p(x)(q(x)ax^n) = (p(x)q(x))ax^n$.

Proof: Fill in the blank.

Theorem: Polynomial multiplication is associative.

Proof: Proceed by induction on the structure or degrees of the three polynomials, using the product ordering. That is, show that assuming the inductive hypothesis:

  • Whenever three polynomials $p(x),q(x),r(x)$ have degrees $d,e,f$, $p(x)(q(x)r(x))=(p(x)q(x))r(x)$

allows you to prove that whenever either:

  • $p$ has degree $d$, $q$ has degree $e$, and $r$ has degree $f+1$,
  • $p$ has degree $d$, $q$ has degree $e+1$, and $r$ has degree $f$, or
  • $p$ has degree $d+1$, $q$ has degree $e$, and $r$ has degree $f$,

then $p(x)(q(x)r(x))=(p(x)q(x))r(x)$.

I'm hoping someone else has a way to improve/flesh out this approach, so I'm making it CW.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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