Show that if a polynomial $f(x)$ in $\mathbb{Z}[x]$ of degree $n$ has no rational root, but for some prime $p$ its reduction mod $p$ has irreducible factors of degrees $1$ and $n - 1$, then $f$ is irreducible.

My attempt: Let $f(x) = a_0 +a_1 x+\ldots+a_n x^n$ and suppose that its reduction modulo $p$, namely $g(x)$, has two factors of degree $1$ and $n-1$. That is, degree of $g(x)$ is $n$, hence $p$ does not divide $a_n$.

$f(x)$ has no rational root, and thus $g(x)$ has no rational root also ( I doubt myself on it). Let $g(x) = g_1(x) g_2(x)$ where degree of $g_1$ is $1$ and degree of $g_2$ is $n-1$. Now $g_1(x) = bx$ because if $g_1(x)$ has a constant term, then $g_1(x)$ has a rational root which leads to $g(x)$ has a rational root, and then $f(x)$ has a rational root also, contradiction.

But I am not sure about it and do not know how to go further.

  • $\begingroup$ There is no particular reason to say $g(x)$ has no rational root, except that strictly speaking the coefficients of $g(x)$ are not integers but residues mod $p$. Indeed you assume $g(x)$ has a factor of degree one, and it would have a root in $\mathbb{Z}/p\mathbb{Z}$ on this account even if the constant term were zero. So I don't see much likelihood of success in pursuing a proof along these lines. $\endgroup$
    – hardmath
    Mar 18, 2015 at 2:42
  • $\begingroup$ How about starting instead from a proof-by-contradiction approach, where you assume $f(x)$ can be factored over $\mathbb{Z}[x]$. What would you be able to say about such a factorization, and how would it relate to the corresponding factorization in $\mathbb{Z}/p\mathbb{Z}$? $\endgroup$
    – hardmath
    Mar 18, 2015 at 2:47

2 Answers 2


Suppose for the sake of contradiction that $f(x)$ is reducible over $\mathbb {Z}[x]$. That is $f(x)=f_1(x) f_2(x)$ for non-units $f_1(x),f_2(x)$.

Then taking residues mod $p$, we would have two factors of $g(x)$ congruent to $f_1(x),f_2(x)$ mod $p$. Since $g_1(x),g_2(x)$ are irreducible (prime) factors of $g(x)$, whose degrees sum to $n$, they must be associates of the reductions mod $p$ of $f_1(x),f_2(x)$.

Since the degrees of $f_1(x),f_2(x)$ also sum to $n$, it follows that their degrees also are $1$ and $n-1$. But having a factor of degree $1$ implies $f(x)$ has a rational root. Contradiction.

  • $\begingroup$ Thanks for the answering. I got it. Let me post my answer and please check it. $\endgroup$
    – user73195
    Mar 19, 2015 at 11:57
  • $\begingroup$ @user73195: I'm not sure if you still wanted me to check your answer. I can see that you deleted it. Perhaps I should say more about how one needs to be careful about the degrees of factors, so that the comparison between what happens over $\mathbb{Z}[x]$ and what over $(\mathbb{Z}/p\mathbb{Z})[x]$ is on solid footing. In general taking the residues mod $p$ might reduce the degree of a factor (it can never increase it), but the sum of degrees being $n$ in both polynomial rings tells us no reduction takes place in this case. $\endgroup$
    – hardmath
    Mar 19, 2015 at 23:58
  • $\begingroup$ Thanks for the help. How to send message to you? $\endgroup$
    – user73195
    Mar 21, 2015 at 12:21
  • $\begingroup$ If you include @hardmath in a Comment (or other post), I will be notified of that post. StackExchange has a chat feature, for which you have enough reputation to participate. The ability to create chat rooms comes by earning additional reputation points. If you let me know in advance, I can arrange to swing by the main Mathematics chat room at a predetermined hour. $\endgroup$
    – hardmath
    Mar 21, 2015 at 16:41
  • $\begingroup$ I don't know how to get the chat feature. :) $\endgroup$
    – user73195
    Mar 21, 2015 at 16:52

Let$f(x) =a_0+a_1x+.....+a_nx^n\in Z[x]$

Suppose $f(x)$ is reducible and $f(x) = g(x) h(x)$ with degree of g, h < n. Further if degree of $g(x)$ is m then degree of $h(x)$ is n-m.
Let $\pi_pf(x)$ is polynomial reduction of $f(x)$ modulo p. $\pi_p f(x) =\pi_pg(x)*\pi_ph(x)$.WLOG degree of $\pi_pg(x)$ is n-1 and degree of $\pi_ph(x)$ is 1.
Hence, if $ h(x)=h_0+h_1x+h_2x^2+.....+h_{n-m}x^{n-m}$, and $\pi_ph(x) = [h_0]+[h_1]x$. That is $h_2,h_3,....,h_{n-m}$ divided by p.

Moreover, the leading coefficient of $f(x)$ is $a_n$, and $\pi_pf(x) $has degree n also. That is $p~is ~not|a_n$.

On the other hand $a_n=h_{n-m}*g_m$ divided by p as shown. Contradiction, hence f(x) is irreducible.

Suppose $h(x) =h_0+h_1(x) $, then $x=\frac {h_0}{h_1}$ is a rational root of f(x) contradiction. Hence f(x) is irreducible.

If $h(x) = h_1x$, then $f(x)$ has no constant term, that leads to $\pi_p(g(x) $has no constant term also, or $[g_0]=0$. But then, $\pi_pg(x)$ is reducible, contradiction.

Hence f(x) is irreducible


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