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Edward Hughes
Frequent poster

Post Number: 180
 Posted on Monday, 19 December, 2011 - 11:50 pm:

Suppose I have A = B.C with B,C in Q[X], A in Z[X], B irreducible. In addition A,B,C monic. According to my notes I should be able to use Gauss' Lemma to show that B,C in Z[X], but I can't see how. Could someone give me a hint. Many thanks, and apologies if I'm being stupid and will see the solution as soon as I step away from the computer!
GCS
Veteran poster

Post Number: 835
 Posted on Tuesday, 20 December, 2011 - 02:58 pm:

There are several variant forms of Gauss's lemma. Which one are you planning to use?
Edward Hughes
Frequent poster

Post Number: 181
 Posted on Thursday, 22 December, 2011 - 04:19 pm:

The one that says that a poly in Z[X] must be irreducible in Q[X] is what I usually refer to as Gauss' Lemma. Any ideas?
George
Veteran poster

Post Number: 469
 Posted on Thursday, 22 December, 2011 - 06:12 pm:

From Wiki i got Gauss' lemma as: A non constant poly in Z[x] is irreducible in Z[x] iff it is both irreducible in Q[x] and primitive in Z.

Maybe it would help to show that A is reducible when viewed as a poly in Q[x], so it's reducible in Z[x] which means there exist poly's B',C' non units in Z[x] with A=B'C'.
Edward Hughes
Frequent poster

Post Number: 183
 Posted on Friday, 23 December, 2011 - 01:25 pm:

But now B irreducible so wlog B | B' in Q[X], but B, B' monic, so by division algorithm in Z[X], B in Z[X], so again by division algorithm C in Z[X]. Is this reasoning correct?
George
Veteran poster

Post Number: 471
 Posted on Friday, 23 December, 2011 - 09:02 pm:

No not quite, the problem is that B', C' could be reducible so if B divides one you can't just say that B'=B. So the most you get is B'=BD for some D in Q[x]. Then you are basically faced with the same question as before.

So instead of saying A=B'C', i'd probably say that as A is reducible in Z[x] we can write $A=A_{1}A_{2}...A_{n}$ with $A_{i}$ irreducible and in Z[x]. By Gauss' lemma irreducible in Z[x] gives us that they are irreducible in Q[x]. Over Q[x] we then have that B divides A_{i} for some i as irreducible implies prime in PID. So $A_{i}=\alpha B$, for some $\alpha$. But $A_{i}$ is irreducible in Q[x] and so $\alpha$ is a unit, moreover B and $A_{i}$ are monic so $\alpha=1$. So B is in Z[x]. Then C must be the product of the other $A_{i}$ so in Z[x].
Edward Hughes
Frequent poster

Post Number: 184
 Posted on Saturday, 24 December, 2011 - 05:25 pm:

Ah yes I see how my logic was flawed. That seems like a nontrivial argument to me at the moment though, so I'll work on seeing these things quicker! Many thanks!
kevinm
Regular poster

Post Number: 54
 Posted on Sunday, 25 December, 2011 - 08:48 pm:

The irreducibility of the polynomial $B$ is irrelevant to the question I think, and it seems to be making you want to factor things in $\mathbb Q[x]$ into irreducibles, which isn't necessary.

One way of saying Gauss' Lemma is the following: if $f \in \mathbb Z[x]$ is given by $f(x) = a_0+a_1x + \ldots + a_nx^n$ then let $co(f)$, the content of $f$, be the greatest common divisor of the $a_i$'s. Gauss's Lemma says that if $f_1,f_2\in \mathbb Z[x]$ the $co(f_1f_2) = co(f_1).co(f_2)$. (A polynomial $f$ with $co(f)=1$ is sometimes called "primitive".) This is the key to the irreducibility statement mentioned above as "Gauss' Lemma".

Now if you have $A=BC$ where $B,C \in \mathbb Q[x]$, then we can write $B = bB'$ and $C=cC'$ where $b,c \in \mathbb Q$ (and positive say, for definiteness) and $B',C' \in \mathbb Z[x]$ with $co(B')=co(C')=1$, by "clearing denominators" and then taking but common factors. Now by Gauss' Lemma we have $c(B'C')=1$ so that since $(bc)B'C'$ is equal to $A$, which is an element of $\mathbb Z[x]$ we must have $bc \in \mathbb Z$. But then $bc= co(A)=1$ since $A$ is monic and so must have $co(A)=1$. Thus $c=b^{-1}$. If $B,C$ are monic however, the only way $B'=b^{-1}B$ can have leading term in $\mathbb Z$ is if $b = 1/n$ for some integer $n$, but the same is true for $c$, and the only way both $b$ and $c$ can be of this form and be inverses of each other is if they are both equal to $1$, hence $B$ and $C$ are in $\mathbb Z[x]$ as required.