Prove polynomial is irreducible?

Solution 1:

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.

Solution 2:

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.