Let $R$ be a UFD and $f(x)\in R[x]$ with $d$ as the content of the polynomial. Let $f(x)=df_1(x).$ Prove that $f_1(x)$ is a primitive polynomial.

Question

Let $R$ be a unique factorization domain and $f(x)$ be a non constant polynomial in $R[x].$ We denote the content of the polynomial as $c(f)=d$ and consequently, write $f(x)=df_1(x).$ Prove that $f_1(x)$ is a primitive polynomial.

The proof given is as follows:

Let $f(x)\in R[x]$ be given where $f(x)$ is a nonconstant polynomial with coefficients $a_0, a_1,..., a_n.$ Let $c$ be a gcd of the $a_i$ for $i =0,1,...,n.$ Then for each $i,$ we have $a_i = cq_i$ for some $q_i\in R.$ By the distributive law, we have $f(x) =df_1(x),$ where no irreducible element in $R$ divides all of the coefficients, $q_1,..., q_n$ of $f_1(x).$ Thus, $f_1(x)$ is a primitive polynomial.

However, I am not getting why no irreducible element in $R$ divides all of the coefficients, $q_1,..., q_n$ of $f_1(x)$ ? I think if it does, then we have a contradiction or something like that. But, I don't understand what it is?

score 1 · Accepted Answer · edited Mar 01 '24 at 15:51

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By the properties of the gcd (valid in any gcd domain), $$\gcd(a_0,a_1,\ldots,a_n)=\gcd(cq_0,cq_1,\ldots cq_n)=c\cdot \gcd(q_0,q_1,\ldots q_n)=\\\gcd(a_0,a_1,\ldots,a_n)\cdot \gcd(q_0,q_1,\ldots q_n)$$

Thus, $\gcd(q_0,q_1,\ldots q_n)=1$ and no irreducible element can divide all $q_i$.

edited Mar 01 '24 at 15:51

Thomas Finley

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answered Mar 01 '24 at 15:45

Julio Puerta

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5
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But how do you know that $\gcd(a_0,a_1,...,a_n)$ is a unit? For if it's not, how did you cancel both sides to get the result? – Thomas Finley Mar 01 '24 at 15:50
$\gcd(a_0,a_1,\ldots, a_n)\neq 0$ and we're in a domain. – Julio Puerta Mar 01 '24 at 16:03
You mean we're in an integral domain, right? – Thomas Finley Mar 01 '24 at 16:05
Yes, of course. – Julio Puerta Mar 01 '24 at 16:05
Thanks! I get it. – Thomas Finley Mar 01 '24 at 16:06
Can you explain how did you write $$\gcd(cq_0,cq_1,\ldots cq_n)=c\cdot \gcd(q_0,q_1,\ldots q_n)$$? – Thomas Finley Mar 01 '24 at 16:22
That is a standard result, see the first answer. – Julio Puerta Mar 01 '24 at 16:24
But that thread seems to be dealing in $\Bbb Z.$ – Thomas Finley Mar 01 '24 at 16:27
Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Mar 01 '24 at 16:27

Let $R$ be a UFD and $f(x)\in R[x]$ with $d$ as the content of the polynomial. Let $f(x)=df_1(x).$ Prove that $f_1(x)$ is a primitive polynomial.

1 Answers1