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Problem: Let $h(t)=c_{2m+1}x^{2m+1}+c_{2m}x^{2m}+ \cdots +c_0$ be a polynomial of odd degree ($\geq 3$) over $\mathbb{Z}$.

Suppose that for some prime $p$, $p$ does not divide the leading coefficient.

$p\mid c_i$ ($m+1 \leq i \leq 2m$)

$p^2\mid c_{i} $ $0 \leq i \leq m$

and $p^3$ does not divide $c_0$.

Show that $h(t)$ is irreducible.

My attempt: I tried to prove using contradiction.

Assume $h(t)=(a_mt^m+ \cdots +a_1t+a_0)(b_nt^{n}+ \cdots + b_1t+b_0)$

Now I split the problem into two cases:

  1. $p^2\mid b_0$ which implies $p$ does not divide $a_0$

and the other case is

  1. $p\mid a_0$ and $p\mid b_0$.

I am being able to prove the result for the first case, since from $c_1=b_0a_1+a_0b_1$ which gives $p|b_1$. Proceeding similarly we get $p|b_n$ which is a contradiction.

Someone help me with the second case, or if there is any other way to do this problem please tell.

user26857
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Subham Jaiswal
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2 Answers2

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Proceed a bit like in the usual proof of Eisenstein's criterion.

Assume contrariwise that $h(t)=g(t)f(t)$ is the product of two factors $$ g(t)=a_\ell x^\ell+\cdots+a_1x+a_0 $$ and $$ f(t)=b_n x^n+\cdots+b_1x+b_0 $$ with $\ell+n=2m+1$.

A plan / extended hints:

  1. We know that $c_{2m+1}=a_\ell b_n$, so neither of the leading coefficients can be divisible by $p$.
  2. So let $r$ be the lowest index $i$ such that $p\nmid a_i$, and $s$ be the lowest index $i$ such that $p\nmid b_i$. Show that this implies that $p\nmid c_{r+s}=\sum_{i+j=r+s}a_ib_j$.
  3. Conclude that we must have $r=\ell$ and $s=n$, so all but the leading terms of both factors must be divisible by $p$.
  4. Show that neither $a_0$ nor $b_0$ can be divisible by $p^2$.
  5. Without loss of generality we can assume that $n<\ell$. Why? Show that $n\le m$.
  6. Using known facts about divisibility of the various coefficients show that $a_0b_n$ is the only term in $c_n=\sum_{i+j=n}a_ib_j$ that is not divisible by $p^2$. Why is this a contradiction?

For full credit: Which steps used the fact that the degree of $h(t)$ is odd?

For extras: Read what Wikipedia says about Newton's polygon in the context of Eisenstein's criterion and your problem. You will see that in this case as well as in the case of Eisenstein's criterion the lower convex envelope consists of a single line with a slope that has denominator $2m+1$. The general facts about Newton's polygon then kick in.

Jyrki Lahtonen
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Really the same as Jyrki's solution, mostly to clarify it to myself.

Consider the polynomial $$h(x) = \sum_{k=0}^n c_k x^k$$

such that $v_p(c_n) = 0$, $v_p(c_i) > 0$ for $i < n$ ( see $p$-adic valuation).

Assume that

$$h(x) = f(x) \cdot g(x)$$

with $f$, $g$ of degrees $k>0$, respectively $l>0$, $k+l=n$. Now $a_k \cdot b_l = c_n$, so $v_p(a_k) = v_p(b_l) = 0$.

Consider now the smallest index $i$, with $v_p(a_i) = 0$, and the smallest index $j$, with $v_p(b_j) = 0$. Right away we get $v_p(c_{i+j}) = 0$. So we must have $i=k$, $j=l$. We conclude that $v_p(a_i) > 0$, for $i< k$, and $v_p(b_j) > 0$, for $j< l$. From here we conclude that $v_p(c_0) = v_p(a_0) + v_p(b_0) \ge 1+1=2$. If we stopped here we had Eisenstein criterion.

Assume that we do have the equality $v_p(c_0) = 2$. Then $v_p(a_0) = v_p(b_0) = 1$.

Assume that the polynomials $f$, $g$ are not of equal degree ( guaranteed to happen if $n=\deg h$ is odd); say $k < l$. Then $$c_k = a_k b_0 + a_{k-1} b_1 + \cdots a_0 b_k$$

and all terms after $a_k b_0$ are divisible by $p^2$. Therefore $v_p(c_k) = 1$. Note that $k < \frac{n}{2}$.

orangeskid
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