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So, this is Hungerford problem 9 on page 166. Here's the problem in full:

Let $f(x) = \sum_{i=o}^n a_i x^i \in \mathbb{Z}[x]$. Suppose that for some $k$, $0 < k < n$, and some prime $p$ such that $p \nmid a_n$, $p \nmid a_k$, and $p \mid a_i$ for all $i = 0, \dots, k-1$, but $p^2 \nmid a_0$. Show that $f$ has a factor of degree at least $k$ that is irreducible in $\mathbb{Z}[x]$.

Here's my progress:

If we construct a new polynomial $g$ where $g$ is $\sum_{i=0}^k a_i x^i$ , we know by Eisenstein's criterion that $g$ is irreducible in $\mathbb{Q}[x]$. If we somehow knew that $g$ was irreducible in $\mathbb{Z}[x]$ too (that would follow if $g$ were primitive), then we could say that the smallest thing we could possibly factor out of $f$ that isn't a unit has to be of degree $k$ and that $g$ must be a factor of it. I tried writing $f$ as $C(f) \cdot f_1$ where $C(f)$ is the content of $f$, so that we'd be dealing with a primitive $f_1$ (and of course, the nice things about $g$ don't go away since $p$ doesn't divide every term), but this doesn't necessarily result in $g$ being primitive (right?)!

So, here's where I'm stuck: how can I show I can even factor $f$? How can I show that $g$ is irreducible in $\mathbb{Z}[x]$, whether by showing it's primitive or by some other means?

Thanks so much, everyone!

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"Here's the problem in full:" You seem to have left quite a bit of it out. – Brandon Carter Dec 7 '11 at 6:17
Is $g$ supposed to be the sum from $0$ to $k$, rather than from $0$ to $n$? – Arturo Magidin Dec 7 '11 at 6:24
@GottfriedLeibniz: The problem was the use of <. It messes up the parser. – Arturo Magidin Dec 7 '11 at 6:28
@Arturo: You are correct; I edited quickly without reading. – Brandon Carter Dec 7 '11 at 6:31
I don't think your approach will work. For one thing, it seems like it would be extremely hard for $g(x)$ to be a factor of $f(x)$! For another, if $f(x)$ is irreducible, then your $g(x)$ will be useless. I don't see how, even if you did know that $g(x)$ is irreducible over $\mathbb{Z}$, you could argue that $g(x)$ is a factor of an irreducible divisor of $f$. For that matter, if that argument held, then why not just factor out the content of $g(x)$ and use the primitive part? No, I don't think this works. – Arturo Magidin Dec 7 '11 at 6:37
up vote 2 down vote accepted

By induction on $n$.

Write $f(x) = C(f) \cdot f_1(x)$ as you have above. Then $f_1(x)$ is primitive. If $f_1$ is irreducible, then we are done. Otherwise, factor $f_1(x) = g(x)h(x)$. If we write $$g(x) = b_0 + b_1x + \dots + b_s x^s$$ $$h(x) = c_0 + c_1 x + \dots + c_t x^t$$ then $b_0 c_0 = a_0$. Since $p \mid a_0$, $p^2 \nmid a_0$, $p$ must divide exactly one of $b_0, c_0$. WLOG, say $p \mid b_0$. We know $p \mid a_1$, but $a_1 = b_0 c_1 + b_1 c_0$. Since $p \mid b_0$ and $p \nmid c_0$, we must have $p \mid b_1$. Continuing inductively, we get $p \mid b_i$ for all $i < k$. If $s < k$, then we get $p \mid g(x)$, which implies that $p \mid f_1(x)$, contradicting that $f_1$ is primitive. Thus $s \geq k$. Hence $g(x)$ is a (primitive) polynomial of degree less than $n$ satisfying the hypothesis, and so has an irreducible factor of degree $\geq k$. But any irreducible factor of $g(x)$ must be an irreducible factor of $f(x)$, so we are done.

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That makes perfect sense. Great answer! Thank you so much! – Reeve Dec 7 '11 at 7:23

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