0

I am reading a textbook titled "Algebra" by Mark R. Sepanski.

In page 179, the author claims that if $D$ is a UFD then every nonzero polynomial $f=\sum_{k} a_k x^k$ (only finitely many nonzero $a_k$'s) in $D[x]$ can be written as a product of an element $c\in D$ and the polynomial $f_0\in D[x]$ which is primitive by taking $c$ to be the gcd of $a_k$'s.

This is how I see it: If $c$ is the gcd of $a_k$'s we have that $a_k=ca_k'$ for some $a_k'\in D$. Now we can consider the polynomial $f_0=\sum a_k'x^k$. So we have $f=cf_0$ and finally we want to prove that $f_0$ is primitive.

I do not see how this could be done without using Bézout's identity which holds in $\mathbb Z$. If Bézout's identity worked for any UFD $D$ then we could write that $\sum_k f_ka_k=c$ for some $f_k$'s in $D$. Then $\sum_k f_k a_k' =1$. That would force gcd of $a_k$'s to be 1.

I am not really sure if Bézout's identity holds in an arbitrary UFD $D$ and I would not expect that as well.

So how do I prove the author's claim without Bézout's identity?

ashK
  • 3,985
  • What's your definition of a primitive polynomial? – Bernard Jun 27 '21 at 10:21
  • @Bernard A polynomial is said to be primitive if gcds of its coefficients are units. – ashK Jun 27 '21 at 10:24
  • Well, if the coefficients of $f$ are divided by their gcd, it seems pretty obvious that the gcd of the quotients is a unit. – Bernard Jun 27 '21 at 10:25
  • @Bernard I fail to see how to prove that. That would mean that I would want to prove that if $c'$ is a gcd of $a_k'$'s then it has an inverse. What would be its inverse? – ashK Jun 27 '21 at 10:33
  • Using the fact that you are working over a UFD, express $c'$ as a product of irreducibles: $c'=u\cdot\prod_{i=1}^rp_i^{m_i}$ for pairwise non associated irreducible $p_1,\dots,p_r$ and positive exponents $m_i$. What you want is that $r=0$. If that weren't the case then the $a_k'$ have a common factor, say $p_1^{m_1}$ which means that $cp_1^{m_1}$ divides all $a_k$ and thus, by definition of $c$, must divide $c$. But then since you are working in a domain, $p_1^{m_1}$ divides $1$ and so $p_1$ is a unit (since $m_1$ is positive by assumption) which contradicts it being irreducible. – Olivier Bégassat Jun 27 '21 at 11:06
  • 1
    You don't even need considering decomposition into irreducibles: if $c' $ is a non-unit dividing all the $a'_k$, then $c'c$ divides all the $a_k$, which contradicts $c$ being $the gcd of the $a_k$. Actually, it seems to bve true, more generally, for gcd domains. – Bernard Jun 27 '21 at 11:14
  • As here, by the linked gcd distributive law: $,d = \gcd(a_1,\ldots,a_n)\Rightarrow \gcd(a_1/d,\ldots,a_n/d) = 1,,$ which is true in any GCD domain, e.g. any UFD. – Bill Dubuque Jun 27 '21 at 14:55

0 Answers0