# primitive, Gauss Lemma

• Jan 18th 2009, 12:19 PM
xianghu21
primitive, Gauss Lemma
Let $\displaystyle A$ be a commutative ring and let $\displaystyle A[x]$ be the ring of polynomials in an indeterminate $\displaystyle x$, with coeﬃcients in $\displaystyle A$. Let $\displaystyle f = a_0 + a_1x + \cdots + a_nx^n \in A[x]$. $\displaystyle f$ is said to be primitive if $\displaystyle (a_0, a_1, \cdots, a_n)=(1)$. Prove that if $\displaystyle f$, $\displaystyle g \in A[x]$, then $\displaystyle fg$ is primitive iff $\displaystyle f$ and $\displaystyle g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\displaystyle \Leftarrow$ which I assume uses Gauss Lemma.

For the $\displaystyle \Rightarrow$, I have:
Suppose that $\displaystyle fg$ is primitive. Now suppose that $\displaystyle f$ were not primitive.
Then $\displaystyle (a_0, a_1, \ldots, a_n) = (1)$ so there is a common factor to all of the terms. But then $\displaystyle fg$ would have a common factor as well. Thus $\displaystyle f$ is primitive, and so must $\displaystyle g$ be.
• Jan 18th 2009, 01:22 PM
ThePerfectHacker
Quote:

Originally Posted by xianghu21
Let $\displaystyle A$ be a commutative ring and let $\displaystyle A[x]$ be the ring of polynomials in an indeterminate $\displaystyle x$, with coeﬃcients in $\displaystyle A$. Let $\displaystyle f = a_0 + a_1x + \cdots + a_nx^n \in A[x]$. $\displaystyle f$ is said to be primitive if $\displaystyle (a_0, a_1, \cdots, a_n)=(1)$. Prove that if $\displaystyle f$, $\displaystyle g \in A[x]$, then $\displaystyle fg$ is primitive iff $\displaystyle f$ and $\displaystyle g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\displaystyle \Leftarrow$ which I assume uses Gauss Lemma.

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

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

Prove the result above. After that, Gauss lemma's follows.
• Jan 18th 2009, 01:25 PM
NonCommAlg
Quote:

Originally Posted by xianghu21
Let $\displaystyle A$ be a commutative ring and let $\displaystyle A[x]$ be the ring of polynomials in an indeterminate $\displaystyle x$, with coeﬃcients in $\displaystyle A$. Let $\displaystyle f = a_0 + a_1x + \cdots + a_nx^n \in A[x]$. $\displaystyle f$ is said to be primitive if $\displaystyle (a_0, a_1, \cdots, a_n)=(1)$. Prove that if $\displaystyle f$, $\displaystyle g \in A[x]$, then $\displaystyle fg$ is primitive iff $\displaystyle f$ and $\displaystyle g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $\displaystyle \Leftarrow$ which I assume uses Gauss Lemma.

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

let $\displaystyle f(x)=\sum_{i=0}^na_ix^i,$ and $\displaystyle g(x)=\sum_{j=0}^mb_jx^j.$ let $\displaystyle f(x)g(x)=\sum_{k=0}^{m+n}c_kx^k.$ first see that by the definition of the coefficients $\displaystyle c_k$ if $\displaystyle t_k \in R, \; 0 \leq k \leq m+n,$ then $\displaystyle \sum_{k=0}^{m+n}t_kc_k=\sum_{i=0}^nr_ia_i=\sum_{j= 0}^ms_jb_j,$ for

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

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

$\displaystyle \overline{f}(x)\overline{g}(x)=(\sum_{i=0}^{n}\ove rline{a_i}x^i)(\sum_{j=0}^{m}\overline{b_j}x^j)=\s um_{k=0}^{m+n}\overline{c_k}x^k=0.$ but $\displaystyle \frac{R}{\mathfrak{m}}[x]$ is an integral domain. thus either $\displaystyle \overline{f}(x)=0$ or $\displaystyle \overline{g}(x)=0,$ which means either $\displaystyle \sum_{i=0}^nRa_i \subseteq \mathfrak{m}$ or $\displaystyle \sum_{j=0}^mRb_j \subseteq \mathfrak{m}.$ contradiction!
• Jan 18th 2009, 02:22 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker

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

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.
• Jan 18th 2009, 07:02 PM
ThePerfectHacker
Quote:

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.
• Jan 23rd 2009, 12:58 AM
GaloisTheory1
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.