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I'm talking about this proof and I don't understand bold part:


Theorem: Let

$f(x)=a_0+a_1x+...+a_nx^n$ be a polynomial with integer coefficients. Suppose a prime $p$ divides each of $a_0,a_1,...,a_{n−1}$ (every coefficient except the leading coefficient), and that $p^2$ does not divide $a_0$. Then $p(x)$ has no factors with integer coefficients.

Proof: Suppose $f=gh$ for polynomials $g,h$ with integer coefficients. Look at this factorization modulo $p$: we get $f(x)=a_nx^n$,

so $g(x)=b_dx^d$, $h(x)=c_ex^e$ for some constants $b_d,c_e$ and for some integers $d,e$ with $de=n$.

This implies the constant term of g(x) is a multiple of p, and similarly for the constant term of h(x), hence $p^2$ divides the constant term of f(x), a contradiction.


Let's call the coefficients of $g$ $b_i$ and of $h$ $c_i$ and suppose $g$ is of degree $r$ and $h$ of degree $s$. Then the coefficient of $x^i$ is $\sum\limits_{\alpha + \beta = i}b_\alpha c_\alpha$. This must be $0$ for $0 \leq i < n$ and $a_n$ for $i = n$. So $b_r c_s = a_n$ but why does this imply $b_\alpha$ and $b_\beta$ must be zero except for $b_r$ and $c_s$?

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  • $\begingroup$ @DonAntonio the constant part of $g$ and $h$ are zero modulo $p$ (or so they claim), so they must be a multiple of $p$. But it's the part before that I don't understand. $\endgroup$
    – user388557
    Commented Mar 12, 2018 at 22:48
  • $\begingroup$ I've now understood the proof...and it is pretty nice, though slightly subtle... $\endgroup$
    – DonAntonio
    Commented Mar 12, 2018 at 22:54

2 Answers 2

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You've already reduced $f$ modulo $p$, a prime, so you have a monomial in a field, $\mathbb{Z}/p\mathbb{Z}$. Consequently, it must factor as a product of monomials having the stated properties of coefficients and degrees.

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  • $\begingroup$ Why must it factor as a product of monomials? $\endgroup$
    – user388557
    Commented Mar 12, 2018 at 22:53
  • $\begingroup$ @PeldePinda : All the roots are zero. And we know something about the number of roots of a polynomial over (sufficient, but unneeded here, extensions) of a field. $\endgroup$ Commented Mar 12, 2018 at 22:57
  • $\begingroup$ $X^2(X^2 + 1)$ has only root $0$, but it can still be factored in non-monomial polynomials. Sorry for being pedantic, but I really want to understand this properly. The multiplicity of the root is equal to the degree of the polynomial, I feel like that is important, but I don't see why. $\endgroup$
    – user388557
    Commented Mar 12, 2018 at 23:11
  • $\begingroup$ @PeldePinda : $X^2(X^2 + 1)$ factors (completely, to linear factors) in an extension field, containing square roots of $-1$. The polynomial $a_n x^n$ immediately factors (completely, to linear factors) in any field and you know all the roots are $0$, so any divisor of that polynomial must have all roots $0$. $\endgroup$ Commented Mar 12, 2018 at 23:16
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Suppose $\;f(x)=a_nx^n+\ldots+a_0\;$, with all its coefficients $\;a_m,\,\,m<n\;$, divisible by $\;p\;$ and $\;a_0\;$ not divisible by $\;p^2\;$. Since reduction modulo $\;p\;$ respects both sums and products of integers, we have that if $\;g(x)=b_dx^d+\ldots+b_0\;,\;\;h(x)=c_ex^e+\ldots+c_0\;$ are such that $\;f(x)=g(x)h(x)\;$ , then reducing modulo $\;p\;$ both sides we get

$$f(x)\pmod p=a_nx^n=(g(x)\pmod p)(h(x)\pmod p)\implies \text{it must be that}$$

$$\;g(x)=\pmod p=b_dx^d\;,\;\;h(x)=c_ex^e\pmod p$$

since otherwise there'd be some powers of $\;x\;$ less than $\;n\;$ in the product $\;g(x)h(x)\pmod\;$ , which is impossible since $\;f(x)=a_nx^n\;$ ...

Thus, both $\;b_0=c_0=0\pmod p\;$ , which then means $\;b_0c_0=a_0\;$ is divisible by $\;p^2\;$ , contradiction...

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  • $\begingroup$ In $(X - 2)(X + 2)$ for example, $X$ to the power of 1 is not in the product because the $-2X$ gets cancelled by $2X$. Isn't it possible that this happens with all powers of $X$? $\endgroup$
    – user388557
    Commented Mar 12, 2018 at 23:13
  • $\begingroup$ @PeldePinda Of course it is possible,yet that's not the main point but rather that the product of the free coeficcients is divisible by $\;p^2\;$ , and here $\;(-2)\cdot3=0\pmod2^2\;$ . Take into account that the free coefficient of $\;f(x)\;$ is the product of the free coefficients of $\;g,\,h\;$ $\endgroup$
    – DonAntonio
    Commented Mar 12, 2018 at 23:18

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