# Prep for Algebra Comprehensive Exam #3

1. Jul 24, 2007

### 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?

2. Jul 24, 2007

### morphism

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

3. Jul 24, 2007

### 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 $$\mathbb{Z}$$?

4. Jul 24, 2007

### morphism

Gauss's lemma.

5. Jul 24, 2007

### 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!

6. Jul 24, 2007

### morphism

No problem.

7. Jul 24, 2007

### 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.

8. Jul 24, 2007

### 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 $$\mathbb{Z}$$.

9. Jul 30, 2007

### mathwonk

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

10. Aug 9, 2007

### BSMSMSTMSPHD

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.

11. Aug 9, 2007

### Kummer

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

12. Aug 9, 2007

### BSMSMSTMSPHD

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

13. Aug 9, 2007

### Kummer

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.