Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an indeterminate $x,$ with coefficients in $A.$ Let $f=a_0 + a_1x+\cdots+a_nx^n \in A[x].$ $f$ is said to be primitive if $(a_0,a_1,\ldots,a_n)=(1).$

Prove that if $f,g\in A[x],$ then $f$ and $g$ are primitive $\Rightarrow$ $fg$ is primitive.

I know Gauss lemma holds true for UFDs, but in this case it isn't required $A$ to be UFD: is it possible to drop the assumption, or is some hypothesis missing here?

  • 3
    $\begingroup$ I am self-studying AM commutative algebra, and I am NOW stuck at the same part of the same exercise. :) $\endgroup$
    – Amr
    Feb 24, 2014 at 9:53
  • $\begingroup$ I read on wikipedia that it is possible to do that if we use the defintion of primitive as in AM's book. I did not look at the proof though. They call this property comaximal instead of primitive $\endgroup$
    – Amr
    Feb 24, 2014 at 9:54
  • $\begingroup$ As a hint (Which I am still trying to make it work up till now): Prove the theorem in the case deg(f),deg(g)=1 $\endgroup$
    – Amr
    Feb 24, 2014 at 9:57
  • $\begingroup$ I believe that main motto of gauss lemma is to prove that $R$ is a U.F.D then so is $R[x]$.. the point of statement product of primitive polynomials is primitive is only intermediate result... So, I suggest you to think of gauss lemma as that and for the matter of primitive polynomial statement you can very well make use of answer given below.... $\endgroup$
    – user87543
    Feb 24, 2014 at 10:41

2 Answers 2


Assume that $fg$ is not primitive. Then the ideal of coefficients of $fg$ is contained in a maximal one, say $\mathfrak m$. In $(A/\mathfrak m)[x]$ we have $f\ne 0$ and $g\ne 0$, so $fg\ne 0$, a contradiction.

  • $\begingroup$ Commutativity is not used in this proof, right? $\endgroup$
    – Atom
    Jan 8 at 13:52

There is an argument which provides explicit $A$-linear combinations equalling $1$, given witnesses to the primitivity of $f$ and $g$ (that is, coefficients $A_0, A_1, \ldots, A_n \in A$ satisfying $A_0a_0 + \cdots + A_na_n = 1$ for $f$, and likewise for $g$).

We proceed by induction on $n+m$. Let the ideal of coefficients of $fg$ be denoted $I_{n,m}$. Let $a_0, \cdots,a_n$ denote the coefficients of $f$, and let $b_0,\cdots,b_m$ denote the coefficients of $g$.

Now, $I_{n,m} = (a_0b_0,a_0b_1 + a_1b_0, \cdots) \subseteq (a_0) + (a_1b_0,a_1b_1 + a_2b_0, \cdots) = (a_0) + I_{n-1,m}$, where $I_{n-1,m}$ is the ideal $(a_1b_0,a_1b_1 + a_2b_0, \cdots)$.

Likewise, $I \subseteq (b_0) + (a_0b_1,a_0b_2 + a_1b_1, \cdots) = (b_0) + I_{n,m-1}$.

Note that $((a_0) + I_{n-1,m})((b_0) + I_{n,m-1}) = ((a_0) + I_{n,m})((b_0) + I_{n,m}) \subseteq I_{n,m}$.

But over the ring $A/(a_0)$, the polynomials $a_1 + a_2x + \cdots + a_m x^{m-1}$ and $g$ are primitive. The ideal $I_{n-1,m}$ is generated by the coefficients of the products of these two polynomials. Thus, by the inductive hypothesis, considering $I_{n-1,m}$ over the ring $A/(a_0)$, we get an $A$-linear combination of elements of $I_{n-1,m}$ equal to $1$ up to an element of $(a_0)$. Similarly we get an $A$-linear combination of elements of $I_{n,m-1}$ equal to $1$ up to an element of $(b_0)$.

Putting the $1$ on one side for each expression, then multiplying them together, gives us what we want. The exact element of $(a_0)$ is found by using the witness of $f$ in the form $A_1a_1 + \cdots + A_na_n = 1 - A_0a_0$ for the inductive step.

The base case is trivial. The calculations get tedious very quickly for larger $n,m$. The expressions found by this argument do not have coefficients of lowest-possible degree in the $a_i$.

Hence Zorn's lemma (for a maximal ideal containing $I_{n,m}$) is not necessary, although that gives the 'quickest' proof.


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