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Given a polynomial $f(x)=c_0+c_1\,x+\ldots+c_n\,x^n\in\mathbb{Z}[x]$, assume that exists $p\in\mathbb{Z}$ prime that satisfies:

  1. $p\,\nmid\,c_n$
  2. $p\,\mid\,c_i,\;\forall i=0,\,\ldots,n-1$.
  3. $p^2\,\nmid\,c_k\; $ for at least one index $k$, $\; 0\leq k\leq n-1$.

Take the minimum $k$ that satisfies (3) and denote it by $k_0$. Suppose that there is a factorization of $f(x)$ in $\mathbb{Z}[x]$:

$$f(x) = g(x)\,h(x).$$

Show that

$$\min\left\{\deg(g(x)), \deg(h(x))\right\}\leq k_0$$


If I consider the projection morphism: $\phi:\,\mathbb{Z}[x] \longrightarrow \mathbb{Z}/p^2[x]$, that sends the coefficients of a polynomial to their class, I see that this morphism preserves the degree of the polynomials $f,\,g$ and $h$, because the class of their leading terms is non-zero. For $\phi(f(x))$, all terms of index less than $k_0$ are zero. I'm looking for applying Eisenstein here and find a contradiction. However, $\mathbb{Z}/p^2$ is not a UFD...


I just want some clue. Because I don't know where to start.

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We write $f = \sum_k c_kx^k$, $g = \sum_k a_kx^k$ and $h = \sum_k b_kx^k$. Working $\bmod p$, the factorization $f = gh$ induces a factorization $\phi(f) = \phi(g) \phi(h)$, where $\phi \colon \mathbb Z[x] \to \mathbb Z/(p)[x]$ denotes the canonical morphism. Now $\phi(f) = c_nx^n$. As $\mathbb Z/(p)[x]$ is a UFD, we must have $\phi(g) = a_mx^m$ and $\phi(h) = b_lx^l$ where $m = \deg g$ and $l = \deg h$. So $p\mid a_k$ for $k < m$ and $p \mid b_k$ for $k < l$. We have \[ c_{k_0} = \sum_{i+j = k_0} a_ib_j \] As $p^2 \nmid c_{k_0}$, we must have $p^2 \nmid a_ib_j$ for some $i$, $j$. But this is only possible if $i \ge m$ or $j \ge l$. That is, \[ k_0 = i+j \ge \max\{i,j\} \ge \max\{m,l\}. \]

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I just needed a clue, but thanks. – synack Oct 22 '12 at 12:11

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