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Prove that if all the roots of a polynomial in $\mathbb{Q}[x]$ are integers, then polynomial is in $\mathbb{Z}[x]$

Efforts:

Let $p(x)=a_0+a_1x +\dots a_nx^n$ be a polynomial in $Q[x]$

We are given that $p(x)$ has all roots in $Z$ so $$p(x)=(x-b_1)(x-b_2)(x-b_3)\dots (x-b_n).$$

Expanding it we get $p(x)=x^n-(\sum b_i )x^{n-1}+(\sum b_ib_j)x^{n-2}+\dots (-1)^nb_1\dots b_n$

Comparing the coefficient we have $a_n=1$, $a_{n-1}=-\sum b_i, \dots, a_0=(-1)^n b_1b_2\dots b_n$ and so on.

Since $b_i$ are integers so is their product. Hence we are done.

Is the proof correct?

Thanks for reading and help!

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    $\begingroup$ How about $\frac12x-\frac12$? $\endgroup$ Feb 6, 2019 at 6:51
  • $\begingroup$ There is something missing when you decompose the polynomial as a function of its integer roots. $\endgroup$ Feb 6, 2019 at 6:53
  • $\begingroup$ @StammeringMathematician Yes, you need the extra hypothesis that the leading coefficient is an integer. $\endgroup$ Feb 6, 2019 at 6:55
  • $\begingroup$ @LordSharktheUnknown Sorry I deleted my comment. So the result is false as stated right now in question. I have to add an extra condition that polynomial is monic. Right? $\endgroup$ Feb 6, 2019 at 6:56

1 Answer 1

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If the given polynomial in $\mathbb{Q}[x]$ is monic, or more generally if its leading coefficient is an integer, then the statement is correct. Note that, in your proof the factorization of $p$ should be $$p(x)=a_n(x-b_1)\dots(x-b_n).$$

Otherwise we have counterexamples: take a monic polynomial with all the roots in $\mathbb{Z}$ and divide it by an integer number greater than $1$.

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  • $\begingroup$ So the result I have stated is true only for monic polynomials? $\endgroup$ Feb 6, 2019 at 6:55
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    $\begingroup$ Note that, given the roots of a polynomial, then its factorization is $a_n(x-b_1)\dots(x-b_n)$. So the result is correct also when the coefficient $a_n$ is an integer. $\endgroup$
    – Robert Z
    Feb 6, 2019 at 6:58
  • $\begingroup$ I really appreciate your quick response and help. Thanks and have a nice day! $\endgroup$ Feb 6, 2019 at 7:03
  • $\begingroup$ It all depends on the leading coefficient. $\endgroup$
    – Wuestenfux
    Feb 6, 2019 at 7:25

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