1. ## primitive, Gauss Lemma

Let $A$ be a commutative 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 $(a_0, a_1, \cdots, a_n)=(1)$. Prove that if $f$, $g \in A[x]$, then $fg$ is primitive iff $f$ and $g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\Leftarrow$ which I assume uses Gauss Lemma.

For the $\Rightarrow$, I have:
Suppose that $fg$ is primitive. Now suppose that $f$ were not primitive.
Then $(a_0, a_1, \ldots, a_n) = (1)$ so there is a common factor to all of the terms. But then $fg$ would have a common factor as well. Thus $f$ is primitive, and so must $g$ be.

2. Originally Posted by xianghu21
Let $A$ be a commutative 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 $(a_0, a_1, \cdots, a_n)=(1)$. Prove that if $f$, $g \in A[x]$, then $fg$ is primitive iff $f$ and $g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\Leftarrow$ which I assume uses Gauss Lemma.

For the $\Rightarrow$, I have:
Suppose that $fg$ is primitive. Now suppose that $f$ were not primitive.
Then $(a_0, a_1, \ldots, a_n) = (1)$ so there is a common factor to all of the terms. But then $fg$ would have a common factor as well. Thus $f$ is primitive, and so must $g$ be.
Let $f(x) = a_nx^n + ... + a_1x + a_0$ and let $g(x) = b_mx^m+...+b_1x+b_0$. Define $h(x) = f(x)g(x)$. Let $\pi$ be an irreducible in $R$. Now if $a_p$ is smallest coefficient not divisible by $\pi$ and $b_q$ is smallest coefficient not dividible by $\pi$ then $c_{p+q}$ is smallest coefficient (in $h(x) = c_Nx^N + ... + c_0$) not divisible by $\pi$.

Prove the result above. After that, Gauss lemma's follows.

3. Originally Posted by xianghu21
Let $A$ be a commutative 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 $(a_0, a_1, \cdots, a_n)=(1)$. Prove that if $f$, $g \in A[x]$, then $fg$ is primitive iff $f$ and $g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\Leftarrow$ which I assume uses Gauss Lemma.

For the $\Rightarrow$, I have:
Suppose that $fg$ is primitive. Now suppose that $f$ were not primitive.
Then $(a_0, a_1, \ldots, a_n) = (1)$ so there is a common factor to all of the terms. But then $fg$ would have a common factor as well. Thus $f$ is primitive, and so must $g$ be.
let $f(x)=\sum_{i=0}^na_ix^i,$ and $g(x)=\sum_{j=0}^mb_jx^j.$ let $f(x)g(x)=\sum_{k=0}^{m+n}c_kx^k.$ first see that by the definition of the coefficients $c_k$ if $t_k \in R, \; 0 \leq k \leq m+n,$ then $\sum_{k=0}^{m+n}t_kc_k=\sum_{i=0}^nr_ia_i=\sum_{j= 0}^ms_jb_j,$ for

some $r_i, s_j \in R.$ so if $fg$ is primitive, clearly both $f$ and $g$ must be primitive too.

conversely, suppose $f$ and $g$ are primitive but $fg$ is not primitive. so $\sum_{k=0}^{m+n}Rc_k \subseteq \mathfrak{m}$, for some maximal ideal $\mathfrak{m}$ of $R.$ now for any $z \in R$ let $z+\mathfrak{m}=\overline{z} \in \frac{R}{\mathfrak{m}}.$ so $\overline{c_k}=0, \; 0 \leq k \leq m+n.$ thus:

$\overline{f}(x)\overline{g}(x)=(\sum_{i=0}^{n}\ove rline{a_i}x^i)(\sum_{j=0}^{m}\overline{b_j}x^j)=\s um_{k=0}^{m+n}\overline{c_k}x^k=0.$ but $\frac{R}{\mathfrak{m}}[x]$ is an integral domain. thus either $\overline{f}(x)=0$ or $\overline{g}(x)=0,$ which means either $\sum_{i=0}^nRa_i \subseteq \mathfrak{m}$ or $\sum_{j=0}^mRb_j \subseteq \mathfrak{m}.$ contradiction!

4. Originally Posted by ThePerfectHacker

Let $f(x) = a_nx^n + ... + a_1x + a_0$ and let $g(x) = b_mx^m+...+b_1x+b_0$. Define $h(x) = f(x)g(x)$. Let $\pi$ be an irreducible in $R$. Now if $a_p$ is smallest coefficient not divisible by $\pi$ and $b_q$ is smallest coefficient not dividible by $\pi$ then $c_{p+q}$ is smallest coefficient (in $h(x) = c_Nx^N + ... + c_0$) not divisible by $\pi$.

Prove the result above. After that, Gauss lemma's follows.
be careful! the ring is not even an integral domain! the problem is not that straightforward.

5. Originally Posted by NonCommAlg
be careful! the ring is not even an integral domain! the problem is not that straightforward.
Yes, you are correct. I usually only deal with unique factorization domains.
I guess this proof can work for special types of domains.

6. Alternatively:

=> contrapositive. wlog suppose f is not primitive. let I be the proper ideal generated by the coefficients of f. It is easily seen that all the coefficients of fg lie in I so that fg is not primitive.

Other direction requires a rigorous argument like above.