2
$\begingroup$

Let $f\in\Bbb{Z}[X]$ be a nonconstant polynomial that is not monic, i.e. its leading coefficient is not $\pm1$. Is is true that $\Bbb{Z}[X]/(f)$ is not finitely generated as an abelian group?

Trying a few simple nonmonic polynomials seems to verify this, and the problem seems to be that you 'get denominators'. I'm not sure how to formalise this however.

For what it's worth, I'm only interested in polynomials with trivial content, i.e. polynomials of which the greatest common divisor of the coefficients equals $1$.

$\endgroup$
6
  • 3
    $\begingroup$ If it is finitely generated then $X\bmod (f)$ is integral over $\mathbb Z$, and thus there is $g\in\mathbb Z[X]$ monic such that $g(X\bmod (f))=0$. This means that $g\in(f)$. Conclusion? $\endgroup$
    – user26857
    Apr 12 '17 at 16:09
  • 1
    $\begingroup$ If $(\mathbb{Z}[X]/(f(X)),+)/\mathbb{Z} = n$ then in $\mathbb{Z}[X]/(f(X))$ : $X^n = h(X)$ with $deg(h) < n$, i.e. in $\mathbb{Z}[X]$ : $X^n = h(X)+ g(X)f(X)$, which is impossible if $f$ isn't monic. $\endgroup$
    – reuns
    Apr 12 '17 at 16:21
  • 1
    $\begingroup$ math.stackexchange.com/questions/903360/… $\endgroup$
    – Viktor Vaughn
    Apr 12 '17 at 16:47
  • 1
    $\begingroup$ Three great answers; anyone care to post an answer so that I can mark this question 'answered'? $\endgroup$
    – user251573
    Apr 12 '17 at 17:37
  • $\begingroup$ Concerning the link; let $\alpha$ denote the image of $X$ in $\Bbb{Z}[X]/(f)$. Then $\alpha$ is not integral over $\Bbb{Z}$, so $\Bbb{Z}[\alpha]\cong\Bbb{Z}[X]/(f)$ is not finitely generated as a $\Bbb{Z}$-module, i.e. as an abelian group. $\endgroup$
    – user251573
    Apr 13 '17 at 17:24
0
$\begingroup$

If $\mathbb{Z}[X]/(f)$ is finitely generated, it is generated by finitely many of the powers of $X$. This means that for $X$ sufficiently large we can write $X^n$ as a linear combination of lower powers of $X$ in $\mathbb{Z}[X]/(f)$, so that the monic polynomial $g(X)=X^n-(\text{lower powers})$ is divisible by $f$. Since the leading coefficient of $f$ must divide the leading coefficient of $g$, that means the leading coefficient of $f$ is $\pm1$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy