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Background

Exercise: Show that every nonzero $f(x)$ in $\Bbb{Q}_{\Bbb{Z}}[x]$ can be written in the form $cx^np_1(x)\cdots p_k(x)$, with $c\in \Bbb{Q},n\geq 0$, and each $p_i(x)$ nonconstant irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$....

Solution: By the unique factorization of $\Bbb{Q}[x]$, $f(x)$ is a product of irreducible. Among these factors, replace each associate of $x$ by $\mathrm{x}$, and replace every $p(x)$ which is not a associate of $x$ by $p(0)^{-1}p(x)$. These adjustments leave a constant term factor, so that $f(x)=cx^np_1(x)\cdots p_k(x)$ where $c\in \Bbb{Q},n\geq 0$ and each $p_i(x)$ is irreducible in $\Bbb{Q}[x]$ with constant term $1$. Each $p_i(x)$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$.

Question

In the exercise question above, the notation $\Bbb{Q}_{\Bbb{Z}}[x]$ denotes the ring $\Bbb{Z}+x\Bbb{Q}[x]$. I am having a difficult time deciphering how to construct the expression. I understand that $\Bbb{Z}+x\Bbb{Q}[x]\subset \Bbb{Q}[x]$. So if I have a polynomial $f(x)\in \Bbb{Z}+x\Bbb{Q}[x]$, with coefficients in $\Bbb{Q}$, and if any of the coefficients are fractions, then i would simply clear the denominators by multiplying the entire polynomials by the product of the denominators, or simply take their LCM.

As a simple example,

let $f(x)=195x^3-1343x^2+1597x+2431$

$=\frac{195}{2431}x^3-\frac{1343}{2431}x^2+\frac{1597}{2431}x+1$

$=(3x-13)(5x-17)(13x+11)$.

So we have $p_1(x)=(3x-13),p_2(x)=(5x-17),p_3(x)=(13x+11)$.

I still did not know where $c,x^n$ come from. I looked at the student solution, and I don't know what the author meant by replace every associate of $x$ by $\mathrm{x}$ and to "replace every $p(x)$ which is not a associate of $x$ by $p(0)^{-1}p(x)$".

In the example above, do I let $c=2431$ and consider say $\frac{1343}{2431}$ as an associate to one of the $x$s in one of the $p_i(x)$ factors?

I think there is an ambiguity in when the question states that "every nonzero $f(x)$ in $\Bbb{Q}_{\Bbb{Z}}[x]$", does the author assume that a polynomial $f(x)$ is of the form with no nonzero constant term, or one with nonzero constant term: $a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0$ with $a_0\in \Bbb{Z}$?

On the other hand, if the question assumes that $f(x)$ is a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_mx^m$, then it let $c=a_m$ and factors out $x^m$ with $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_mx^m={a_m}{x^m}(x^{n-m}+\frac{a_{n-1}}{a_m}x^{n-m-1}+\cdots+\frac{a_{m-1}}{a_m}x^2+1)$. Since $x^{n-m}+\frac{a_{n-1}}{a_m}x^{n-m-1}+\cdots+\frac{a_{m-1}}{a_m}x^2+1\in \Bbb{Q}[x]$, the expression factors into irreducibles in the form of $p_1(x)\cdots p_k(x)$.

Thank you in advance

Seth
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  • Factor $f$ into irred's in $\Bbb Q[x].,$ Factors $p$ not a power of $x$ can normalized to have constant coef $= 1$ by dividing $p$ by its lead coef $p(0),,$ balanced by multiplying $c$ by $p(0),,$ i.e. wlog factors are monic (hence $\in\Bbb Q_{\Bbb Z}[x]),,$ by absorbing lead coefs into $c.\ \ $ – Bill Dubuque May 04 '25 at 19:12
  • @BillDubuque I edited my post. I think what you stated is what I did in my example? Except I don't know what you meant by looking at the factors $p$ that are not a power of $x$. Also, when you say multiply $c$ by $p(0)$, is $c$ the constant term of the polynomial $f$? – Seth May 04 '25 at 19:16
  • @BillDubuque so if we have as an example the polynomial $g(x)=x(3x-13)(5x-17)(13x+11)$, with $p_1(x)=x, p_2(x)=(3x-13), p_3(x)=(5x-17),p_4(x)=(13x+11)$, then $p_1(x)$ is call a monic irreducible and $p(0)=0$ is according to the solution an associate of $x$? – Seth May 04 '25 at 19:25
  • @BillDubuque in a degree 3 polynomial as above or the degree 4 one in my previous comment, the least degree coeffs would be the term with degree 1? Also I don't know what is meant by monic as a power series and I am guessing you mean reverse monic as in the usual meaning of monic polynomial where the leading term of a polynomial with coefficient 1? – Seth May 04 '25 at 19:29
  • I meant "reverse monic" above, i.e. monic as a power series, i..e. least degree term has coef $= 1.,$ The idea is to factor out all least degree coefs into $c,$ leaving least degree coef $= 1,,$ so $p(0) !=! 1$ (or $,0,$ if $p(x)!=!x).,$ It's just the reverse of the standard unit normalization of prime factorizations in $,\Bbb Q[x],,$ where wlog we assume all prime factors are monic by cancelling and collecting all lead coef's into a constant factor. $\ \ $ – Bill Dubuque May 04 '25 at 19:41
  • @BillDubuque I edited my post, I think there is a bit of a confusion with the wording of the question. I am not sure what form the author assume $f(x)$ is suppose to take,. I probably am not clear on the wording when it says that "every nonzero $f(x)\in \Bbb{Q}_{\Bbb{Z}}[x]$. does the author mean a polynomial with no nonzero constant terms, or ones that allow for integer constant terms. – Seth May 06 '25 at 00:01
  • Factor $f$ in $\Bbb Q(x)$ then normalize each irred factor $p(x) \neq x$ to be reverse-monic $(p(0)=1)$ by moving a factor $p(0)$ from $p(x)$ into the constant $,c,,$ e.g. $$,\underbrace{c, x^n:! p_1(x), p_2(x)}{{\large \rm factorization\ in}\ \Bbb Q[x]}\ \to\ \underbrace{c,p_1(0),p_2(0), x^n, \frac{p_i(x)}{p_1(0)}, \frac{p_2(x)}{p_2(0)}}{\large {\rm factorization\ in}\ \Bbb Q_{\Bbb Z}[x]}\qquad$$ – Bill Dubuque May 06 '25 at 01:28
  • @BillDubuque can I just ask you one last quick thing. In the quoted solution, when the author says associates of $x$, he means only for monomial terms of the form $\frac{a}{b}x^n$, $\frac{a}{b}\in \Bbb{Q}$, $n\in \Bbb{Z}_{\geq 0}$ and I replace such entire term with $\mathrm{x}$? Am I interpreting that correctly? – Seth May 09 '25 at 23:02
  • By definition $a$ and $b$ are associate if they divide each other $,a\mid b\mid a,,$ so in a domain $,a = ub,$ for $,u,$ a unit (invertible). So the associates of $,x,$ in $,\Bbb Q[x],$ are $,q:!x,, q\in\Bbb Q$. The author collects them all into $,x^n,,$ and collects their coefs $,q_i,$ into $,c,,$ i.e. normalize to $1$ the coef of the least degree term - which amounts to the same reverse-monic normalization as done for all $,p_i(x).\ \ $ – Bill Dubuque May 09 '25 at 23:17
  • @BillDubuque I will be taking my abstract algebra course next year and covers first 14 chapters of Dummit and Foote. If this was a homework question, I don't think neither the TA nor the course professor will expect my answer to be as slick nor as terse as yours. Also, I am going through D&F's text exercise by exercise, and I don't think in any of my written up solutions on my own are as slick as yours. Maybe one day, I will arrive there. Also, I think $c$ includes terms of the form $p(0)^{-1}$. – Seth May 09 '25 at 23:29

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