4
$\begingroup$

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

Let $f(x) = \sum_{i=0}^n a_i x^i \in \mathbb{Z}[x]$. 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!

  • 0
    @Brandon: "Content" is correct. The content is the gcd of the coefficients; "primitive" means that the content is $1$. And $f_1$ is better than $f'$ here, lest it be confused with the (formal) derivative.2011-12-07
  • 0
    How do I change the prime into a subscript 1? I originally wrote it as subscript 1, but one of the mods changed my submission, and that ended up lost in translation (that might explain the k turning into an n, too).2011-12-07
  • 1
    @Arturo: You are correct; I edited quickly without reading.2011-12-07
  • 0
    @Reeve: The $k$ did not turn into $n$: I was very careful to write *exactly* what you wrote there, and *you* wrote $n$, not $k$: check the edit history by clicking on the "Edited xxx ago" link. As for the prime, it's already been changed into a subscript. Refresh the page.2011-12-07
  • 1
    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.2011-12-07
  • 0
    @ArturoMagidin I'm sorry I blamed it on the editing. I checked how I originally had it typed, and you are indeed correct. We're trying to show f(x) is reducible, right? So wouldn't that mean that our g(x) would be useful? And, I agree that it would be _extremely_ hard for g(x) to be a factor; I just realized that we don't need it to be. All we get from the information I deduced is that there is no factor of smaller degree than g, right? Edit: Wait a sec... $x^3 + x^2 + x + 1$ has a factor x+1 but $x^2 + x + 1$ is irreducible. I'm a dummy...2011-12-07
  • 0
    @Reeve: No, you are not trying to show that $f(x)$ is reducible. You are trying to show that it has an irreducible factor of degree at least $k$. If $f(x)$ were itself irreducible over $\mathbb{Q}$, then the primitive part of $f(x)$ would be a witness to the fact that $f(x)$ has an irreducible factor of degree at least $k$. And, in principle, there *could* be irreducible factors of degree *smaller* than $k$, you just need to prove that there is at least one of degree at least $k$.2011-12-07
  • 0
    @ArturoMagidin Okay. I think I'm starting to get what you're saying. I'll think on it some more. So, scrap my initial thoughts because they're garbage?2011-12-07
  • 0
    @Reeve: I just don't see how looking at the truncated polynomial (which is what your $g(x)$ is) is going to help.2011-12-07
  • 0
    Good point. I just don't see how I can apply the Eisenstein criterion otherwise. :/2011-12-07

1 Answers 1

3

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.

  • 0
    That makes perfect sense. Great answer! Thank you so much!2011-12-07