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Edward Hughes
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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!
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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
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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?
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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
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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?
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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 [inline]A=A_{1}A_{2}...A_{n}[/inline] with [inline]A_{i}[/inline] 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 [inline]A_{i}=\alpha B[/inline], for some [inline]\alpha[/inline]. But [inline]A_{i}[/inline] is irreducible in Q[x] and so [inline]\alpha[/inline] is a unit, moreover B and [inline]A_{i}[/inline] are monic so [inline]\alpha=1[/inline]. So B is in Z[x]. Then C must be the product of the other [inline]A_{i}[/inline] 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!
Regular poster

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

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

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

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

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