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Let $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ be polynomial whose coefficients are integers and whose leading coefficient is $1$. If $c$ is integer and root of $f$, show that $c$ divides $a_0$.

(From : Serge Lang - Basic Mathematics, Exercise 5, page 328)

What I tried:


If $c$ is root then:
$f(c)=0$ it follows that,
$f(c)=c^n+a_{n-1}c^{n-1}+...+c+a_0$
$0-a_0=c(c^{n-1}+a_{n-1}c^{n-2}+...+1)$

Let $S=(c^{n-1}+a_{n-1}c^{n-2}+...+1)$
We know that $c$ and the coefficients $a_{n-1},a_{n-2},...,a_0$ are integers, hence S will also be an integer.

It follows that,
$-a_0=cS$

Hence, $c$ divides $a_0$.


Is there a mistake?

Supposing that I didn't made a mistake, it will be the same even if the leading coefficient is not $1$. Is that right?

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  • $\begingroup$ That looks good to me. $\endgroup$ – Xander Henderson Oct 15 '17 at 0:12
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    $\begingroup$ It's quite fine. In the same way, you can prove the full rational roots theorem : if a rational number $a/b$ (in irreducible form) is a root, then the numerator divides the constant term, and the denominator divides the leading coefficient. $\endgroup$ – Bernard Oct 15 '17 at 0:14
  • $\begingroup$ @Bernard If we consider the rational roots theorem, do we have a special case when the leading coefficient is 1? Thanks for the quick replies! $\endgroup$ – Rumata Oct 15 '17 at 0:41
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    $\begingroup$ You can also look at Vieta's formula for the polynomial. What would be $a_0$ in terms of the roots of the polynomial and / or the independent terms of the factorization? $\endgroup$ – Leo Lerena Oct 15 '17 at 0:43
  • $\begingroup$ Well, it yields in this case a rational root must be an integer. $\endgroup$ – Bernard Oct 15 '17 at 1:00

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