Prove polynomial is irreducible?

Question

Suppose we have a monic polynomial $p(x)\in \mathbb{Z}[x]$ of degree greater than 1 with all even integer coefficients except for the leading term such that the coefficient on the linear term is an odd multiple of 2. We know that the polynomial does not have an integer root. How then can we prove that the polynomial is irreducible over the integers?

So far, I've proven that not having an integer root means that the polynomial does not have a root in the rationals either by using the fact that it's monic and reproving the rational root theorem in this case, but I don't see how to use this information to prove that the polynomial is completely irreducible over the integers.

Answer 1

I should probably mention the Gauss Lemma, monic and irreducible over the integers implies irreducible over the rationals.

Begin with the simplest cases, see how that goes. Here, degree $4,$ and we know there are no linear factors, so we are asking whether $$ (x^2 + a x + b)(x^2 + c x + d) $$ can work.

From $x^3$ we get $a+c \equiv 0 \pmod 2.$ So, both odd or both even.

From $x^2$ we get $ac + b+d $ even.

case(I) $a,c$ odd. Then $b+d$ also odd. From the constant term, we know $bd$ is even. Perhaps by renaming, take $b$ even and $d$ odd. However, the coefficient of $x$ is $ad+bc$ which is odd in this case.

case(II) $a,c$ even. Then $b+d$ is even. Since $bd$ must be even, this says both $b,d$ even. In this case, all four letters are even. The linear coefficient is still $ad + bc,$ but with all four letters even we know this is divisible by $4.$

Next try degrees 5 and 6, where 6 has two types, either quadratic times quartic or two cubics.

Answer 2

HINT.-$$p(x)=x^n+2n_1x^{n-1}+2n_2x^{n-2}_2+\cdots+2(2m+1)x+2a_n\\p(x)=x(x^{n-1}+2n_1x^{n-2}+2n_2x^{n-3}_2+\cdots+2(2m+1))+2a_n$$ Then $$p(x)=xq(x)+2a_n$$ where $q(x)$ is irreducible by Einsenstein so the most general form of $q(x)$ is $$q(x)=\prod_ j(x-r_j)\prod(x-(s_k+it_k))(x-(s_k-it_k))=\prod_j(x-r_j)\prod_k(x^2-2s_kx+(s_k^2+t_k^2))$$ where each $r_j$ is irrational real.

It follows that if $p(x_0)=0$ then $$x_0\prod_j(x_0-r_j)\prod_k(x_0^2-2s_kx_0+(s_k^2+t_k^2))+2a_n=0$$ If $a_n=0$ then $p(x)$ has the root $x=0$ which is contrary to the problem ($p(x)$ has no integer root) so we assume that $a_n\ne0$. But in this case $x_0$ is irrational because $q(x_0)\ne0$. We are done.

Answer 3

To clear up some confusion, the claim is:

If $$p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0 \in \mathbb{Z}[x],$$ with $2\mid a_0,a_1,\dots,a_{n-1}$, $4 \nmid a_1$ and $p(x)$ has no integer root, then $p(x)$ is irreducible.

Following generalization of this claim is proven in article A mild generalization of Eisenstein’s criterion by Steven H. Weintraub (see copy at https://www.lehigh.edu/~shw2/preprints/eisenstein.pdf):

Theorem. Let $f(x)=a_nx^n+\dots+a_0 \in \mathbb{Z}[x]$ be a polynomial and suppose there is a prime $p$ such that $p$ does not divide $a_n$, $p$ divides $a_i$ for $i=0,\dots,n-1$, and for some $k$ with $0 \leq k \leq n-1$, $p^2$ does not divide $a_k$. Let $k_0$ be the smallest such value of $k$. If $f(x)=g(x)h(x)$, a factorization in $\mathbb{Z}[x]$, then $\min (\deg(g(x)),\deg(h(x))) \leq k_0$.

Proof. Suppose we have a factorization $f(x)=g(x)h(x)$. Let $g(x)$ have degree $d_0$ and $h(x)$ have degree $e_0$. Let $d$ be the smallest power of $x$ whose coefficient in $g(x)$ is not divisible by $p$, and similarly for $e$ and $h(x)$. Then $g(x)=x^dg_1(x)+pg_2(x)$ and $h(x)=x^eh_1(x)+ph_2(x)$ for polynomials $g_1(x),g_2(x),h_1(x),h_2(x)\in \mathbb{Z}[x]$, with the constant terms of $g_1(x)$ and $h_1(x)$ not divisible by $p$. Then $$ f(x)=g(x)h(x)=x^{d+e}g_1(x)h_1(x)+p(x^eh_1(x)g_2(x)+x^dh_2(x)g_1(x))\\+p^2g_2(x)h_2(x). $$ The condition that all of the coefficients of $f(x)$ except $a_n$ be divisible by $p$ forces $d+e=n$ and hence $d=d_0$ and $e=e_0$. Thus $g(x)=b_{d_0}x^{d_0}+pg_2(x)$ and $h(x)=c_{e_0}x^{e_0}+ph_2(x)$, in which case $$ f(x)=g(x)h(x)=a_nx^n+ph_2(x)b_{d_0}x^{d_0}+pg_2(x)c_{e_0}x^{e_0}\\+p^2g_2(x)h_2(x), $$ and so $k_0\geq \min(d_0,e_0)$. $\square$

In your case $p=2$ and $k_0=1$ (if smallest $k$ with $p^2\nmid a_k$ was $k=0$, you can use Eisenstein criterion directly, hence we can assume $k_0=1$). Then the statement says that if the polynomial was reducible, it must have at least one of its factors with degree $\leq k_0=1$. It cannot be a degree $0$ since your polynomial is primitive, so it must have degree one factor. Since $p(x)$ is monic, this implies an integer root, a contradiction. So $p(x)$ is irreducible.

Answer 4

I used (an instance of) this as a problem in a mid-term 3½ weeks ago, so I guess I might as well.

By Gauss's lemma an eventual factorization over $\Bbb{Q}$ consists of polynomials with integer coefficients. Anyway, let's assume contrariwise that $p(x)=g(x)h(x)$ non-trivially. WLOG $g(x),h(x)$ are monic and have integer coefficients. We can reduce this modulo two, and end up with a factorization in $\Bbb{Z}_2[x]$: $$ \overline{p}(x)=\overline{g}(x)\overline{h}(x). $$ It is given that apart from the leading $1$, the coefficients of $\overline{p}(x)$ all vanish, so $\overline{p}(x)=x^n$.

One of the key observations is that $\Bbb{Z}_2[x]$ is a unique factorization domain as a polynomial ring over a field, so we can conclude that $$\overline{g}(x)=x^m,\quad\overline{h}(x)=x^t$$ with $m+t=n$.

Another key observation is that as $p(x)$ is known to have no integer roots, hence by the rational root test no linear factors, we must have $m\ge2$ as well as $t\ge2$.

Drums, please. It follows that $$ g(x)=x^m+\cdots+g_1x+g_0,\qquad h(x)=x^t+\cdots+h_1x+h_0 $$ with $g_1,g_0,h_0,h_1$ all even integers (as they reduce to zero mod $2$). Consequently the linear term coefficient $p_1$ of $p(x)=x^n+\cdots+p_1x+p_0$ $$ p_1=g_0h_1+g_1h_0 $$ is divisible by four. A contradiction.