Irreducible Polys--Please help/check me

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July 12th 2006, 06:41 AM
harold

Irreducible Polys--Please help/check me

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July 12th 2006, 08:54 AM
ThePerfectHacker

Quote:

Originally Posted by harold

If I know the above (assuming what I wrote above is correct--please check me!!) and the following facts, then how do you conclude irreducibility over $R$ and $C$ ?

It seems to me your work is correct.
I think you are asking how do you determine whether a polynomial is irreducible over $\mathbb{R}[x]$ and $\mathbb{C}[x]$ ?
---
Let me begin by answering the question about $\mathbb{C}[x]$ . Any non-linear polynomial is reducible over $\mathbb{C}$ ! That is called the "fundamental theorem of algebra".

To answer the other question whether a polynomial is reducible over $\mathbb{R}$ . All you need to find its roots. And confirm that they are not complex but real.
July 12th 2006, 10:18 AM
ThePerfectHacker

Quote:

Originally Posted by harold

Yes, that's what I meant. :)

So for instance, for 1):
$x^4 - 15x^2 - 76 = (x-sqrt{19})(x+sqrt{19})(x-2i)(x+2i)$
This is 1) written as a product of irreducibles in $C[x]$ . $(x-2i)$ and $(x+2i)$ are not in $R[x]$ , but they are conjugates, so $(x-2i)(x+2i)=x^2+4$ is in $R[x]$ and is irreducible in $R[x]$

If this is correct, how do I approach 2)--4) ?? How can you factor them like this? This one just happened to be all there; I find that the others are a bit tricky...

Yes,
You have show that $x^4-15x^2-76$ is reduccible over $\mathbb{R}$ such as,
$(x^2+4)(x^2-19)$ . Now can this be reduced further? Well,
$x^2-19=(x-\sqrt{19})(x+\sqrt{19})$
but,
$x^2+4$ cannot be.
July 13th 2006, 09:40 AM
ThePerfectHacker

Quote:

Originally Posted by harold

But how do I formally say there that a--d are irreducible over R and C??

Again you cannot say irreducible over $\mathbb{C}$ except in a one degree polynomial. You want to how to be formally show that they are irreducible over $\mathbb{R}$ .
What you can do, is reducible (factor) the polynomials in $\mathbb{C}$ . Then since, $\mathbb{R} \leq \mathbb{C}$ you can say that since factorization is unique in irreducible polynomials that polynomial (the one you are trying to show is irreducible over the real field) cannot be factored in any other way (namely in real numbers), since the reals are a subfield of the complex numbers.
July 13th 2006, 06:00 PM
BubbleBrain_103

Looking first at what you have so far (part I):

1) is incorrect... you are using the fallacy that simply because a poly has no roots in Q it is necessarily irreducible over Q. This is valid if you are dealing with degree 3 or smaller, but otherwise you could have quadratic or higher factors, which is the case here.. it factors as (x^2+4)(x^2-19), both rational polys, and THESE are irreducible because they have no rational roots (any further factorization would yield a linear factor, hence a rational root)

For the rest, I did not do the criteria calculations or check your factorizations, but assuming they are correct, I find no obvious errors of reasoning.

Part II:

I'm sure I'm repeating here a lot of stuff Hacker already said, but for the sake of continuity I'll say it again...

No non-linear polynomial is irreducible over C. That's the beauty of the complex's, every polynomial completely factors into linear factors. So if it's not linear, keep going.

No polynomial (in real co-efficients) of degree greater than or equal to 3 is irreducible in R. That is, the only irreducibles in R are linear polynomials, and quadratic polynomials having no real root. How do we know this? It follows from the fact that complex roots to polynomials in real coefficients always come in conjugate pairs (not necessarily true if you're allowed COMPLEX coefficients, which doesn't apply here).

As an illustration, say you handed me a third degree polynomial and claimed it was irreducible in R. Then I say, "well, it has a complex root, find it." You give it to me, call it z. Then I also know that its conjugate z' is also a root. So the polynomials (x-z) and (x-z') are factors in your polynomial. Multiply them together, and

(x-z)(x-z') = x^2 -(z+z')x +zz'

is a factor. But (z+z'), zz' are both real numbers. So you have a real quadratic factor.

Moral of the story, if your polynomial is of degree 3 or higher, keep going, it factors further in the reals.

So, in fact, once you've completely factored something over the reals/complexes, its much easier to argue the irreducibility of the factors then, say, over the rationals. In the complexes, they're all going to be linear, obviously irreducible. And for the reals, if you have a quadratic factor, just show that it has no real roots by the quadratic formula.
July 13th 2006, 07:04 PM
ThePerfectHacker

Quote:

Originally Posted by harold

Thanks, BB! I believe I can do it now! Were you able to find the factors for $x^4 - 15x^2 - 75$ . I found the roots which are above (hard to read a little b/c of the latex). This one is tough! :confused:

Let $y=x^2$
Thus, you end up with a quadradic,
$y^2-15y-75$
July 15th 2006, 11:12 PM
malaygoel

Quote:

Originally Posted by harold

What I mean to say in the above is how do I form b) into product of linear factors?

You are on the right way(I think you are asking for this)
$x^4 - 15x^2 - 75$
$=(x^2 - \frac{15 + \sqrt{525}}{2})(x^2 - \frac{15 - \sqrt{525}}{2})$
$=(x^2 - \frac{15 + \sqrt{525}}{2})(x^2 + \frac{\sqrt{525} - 15}{2})$
$=(x-\sqrt{\frac{15 + \sqrt{525}}{2}})$ $(x+\sqrt{\frac{15 + \sqrt{525}}{2}})$ $(x-i\sqrt{\frac{\sqrt{525}-15}{2}})$ $(x+i\sqrt{\frac{\sqrt{525}-15}{2}})$

Keep Smiling
Malay
July 17th 2006, 08:04 PM
malaygoel

Quote:

Originally Posted by harold

How come the last two factors flipped to become ${\sqrt{525}-15}$

I have three steps in my last post. Watch the second step.

Quote:

Also, now that $x^4 - 15x^2 - 75$ is broken down, am I right in saying the following when deducing irreducibility:

$(x-\sqrt{\frac{15 + \sqrt{525}}{2}})$ $(x+\sqrt{\frac{15 + \sqrt{525}}{2}})$ $(x-i\sqrt{\frac{\sqrt{525}-15}{2}})$ $(x+i\sqrt{\frac{\sqrt{525}-15}{2}})$
is written as a product of irreducibles in $\mathbb C[x]$ . (So its irreducible over $\mathbb C[x]$ )

As discussed earlier, each of the four terms is irreducible

Quote:

Since $(x-i\sqrt{\frac{\sqrt{525}-15}{2}})$ and $(x+i\sqrt{\frac{\sqrt{525}-15}{2}})$ are both conjugate pairs that are not in $\mathbb R[x]$ but $(x-i\sqrt{\frac{\sqrt{525}-15}{2}})$ $(x+i\sqrt{\frac{\sqrt{525}-15}{2}}) =$ some poly (I need help finding this poly) which is in $\mathbb R[x]$ and so is irreducible in $\mathbb R[x]$ .

The poly you want to find is the second term in the first step.

Quote:

Am I saying it right when concluding irreducibility? Are these the right steps?

I am referring to the previous post. First you factorised into two quadratic terms, one is reducible in R(x) and the other is irreducible in R(x) but reducible in C(x).After further factorisation, you get four linear terms which are irreducible in C(x).
Please someone check it. I am not very confident.

Keep Smiling
Malay