# Exercise from Atiyah-Macdonald, Chapter 1, 2.iv)

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?

• I am self-studying AM commutative algebra, and I am NOW stuck at the same part of the same exercise. :)
– Amr
Feb 24, 2014 at 9:53
• 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
– Amr
Feb 24, 2014 at 9:54
• 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
– Amr
Feb 24, 2014 at 9:57
• 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....
– user87543
Feb 24, 2014 at 10:41

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.

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.