Prep for Algebra Comprehensive Exam #3

[BSMSMSTMSPHD]

3. Show that the polynomial p(x) = x^4 + 5x^2 + 3x + 2 is irreducible in \bbmath{Q} [x].

This on has me totally stumped. I've seen many other problems on polynomial reducibility, but they are all solvable by Eisenstein, or by checking for irreducibility in a quotient ring by a proper ideal. Neither applies here.

I know that if it does reduce, then there are only two cases to check.

Case 1: p(x) = a(x)b(x) where deg(a(x)) = 3, deg(b(x)) = 1.

This case is easy to eliminate, since it implies that p(x) must have a root in \mathbb{Q}. Since the only possible choices are \pm 1, \pm 2, it is easy to check that none of these work.

So, Case 1 is impossible.

Therefore, if p(x) is reducible, then it must reduce into two monic polynomials of degree 2.

Case 2: p(x) = a(x)b(x) where deg(a(x)) = deg(b(x)) = 2.

Here is where I hit a dead end. I've tried my entire bag of tricks and I have nothing to show for it. Can someone get me started?

 

[morphism]

Suppose it factors into (x^2 + ax + b)(x^2 + cx + d) over \mathbb{Z}. Multiply out, equate coefficients, and derive a contradiction.

 

[BSMSMSTMSPHD]

I did that - but I did it over \mathbb{Q} which made it impossible to find a contradiction (or at least difficult enough that I gave up). Why am I allowed to restrict my coefficients to \mathbb{Z}?

Thanks for your quick response!

 

[morphism]

Gauss's lemma.

 

[BSMSMSTMSPHD]

So, if I'm reading this correctly, Gauss's Lemma says that if R is a UFD (Z in our case) and F is its field of fractions (Q in our case), then a polynomial p(x) in R[x] (which ours is) that is reducible in F[x] is also reducible in R[x].

So I'm taking the contrapositive stance, right? If p(x) is irreducible in Z[x], then it is irreducible in Q[x].

I see here a corollary which I think I can use: Same R and F - if the g.c.d. of the coefficients of p(x) is 1 then p(x) is irreducible in R[x] if and only if it is irreducible in F[x].

Thanks for pointing me in the right direction!

 

[morphism]

No problem.

 

[Kummer]

We can state Gauss' Lemma a little stronger than that.

Theorem: Let D be a UFD and F be the field of quotients of D. Given a polynomial f(x) \in D[x]. The polynomial f(x) factors into degrees n\mbox{ and }m in D[x] if and only if f(x) factors into polynomials of degrees n \mbox{ and }m in F[x].

Since \mathbb{Q} is the field of quotients of \mathbb{Z} Gauss' Lemma applies. We can restrict our attention to the integers because they are easier to work with.

 

[BSMSMSTMSPHD]

Okay, I think I've got it now. Equating (x^2 + ax + b)(x^2 + cx + d) to p(x), we arrive at the following system of equations:

a + c = 0

b + ac + d = 5

ad + bc = 3

bd = 2

From the first equation, I know that c = -a. Substituting into the second equation and moving a few terms gives me a^2 = (b + d) - 5. (*)

Now, from the bottom equation, we know b, d \in \{ \pm 1, \pm 2 \}. But, once we know b, we automatically know d. Looking back at (*) the only possible values for b + d are -3 or 3. Both produce a contradiction, however, since a^2 = -2 and a^2 = -8 have no solutions in \mathbb{Z}.

 

[mathwonk]

mod 3 this reduces to an irreducible polynomial of degree 4. done.

 

[BSMSMSTMSPHD]

mod 3 this reduces to an irreducible polynomial of degree 4. done.

I think I understand what you're saying. Since I can check for irreducibility in \mathbb{Z} I can use the proper ideal 3 \mathbb{Z} to show irreducibilty by Eisenstein.

There is a second part to this question that I did not post earlier. I'll post it now (I've been working on papers for the past 2 weeks...):

Show that the same polynomial does not have any roots in \mathbb{Q} (\sqrt[5]{2}).

Is this just a matter of comparing orders of extensions? If k is a root of the polynomial, then [ \mathbb{Q} (k) : \mathbb{Q} ] = 4 which does not divide [ \mathbb{Q} (\sqrt[5]{2}) : \mathbb{Q} ] = 5.

It seems like the answer to this question is simple, but I'm having a hard time justifying it in the right language.

 

[Kummer]

Theorem: Let E be an extension field over F if \alpha is algebraic over F and \beta \in F(\alpha) then \deg \left<\beta, F\right> \mbox{ divides }\deg \left< \alpha , F \right>.

 

[BSMSMSTMSPHD]

You define the extension E, but then never use it in your statement. I'm confused... Is my explanation correct?

 

[Kummer]

You define the extension E, but then never use it in your statement. I'm confused... Is my explanation correct?

The degree of k is 4 since the polynomial is irreducible over F. Now if k is in Q(2^{1/5}) the by theorem deg(k,Q) divides deg (2^{1/5},Q). But deg (2^{1/5},Q)=5 and so we have that 4 divides 5 a contradiction.