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- Jul 12th 2006, 06:41 AMharoldIrreducible Polys--Please help/check me
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- Jul 12th 2006, 08:54 AMThePerfectHackerQuote:

Originally Posted by**harold**

I think you are asking how do you determine whether a polynomial is irreducible over $\displaystyle \mathbb{R}[x]$ and $\displaystyle \mathbb{C}[x]$?

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Let me begin by answering the question about $\displaystyle \mathbb{C}[x]$. Any non-linear polynomial is reducible over $\displaystyle \mathbb{C}$! That is called the "fundamental theorem of algebra".

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

Originally Posted by**harold**

You have show that $\displaystyle x^4-15x^2-76$ is reduccible over $\displaystyle \mathbb{R}$ such as,

$\displaystyle (x^2+4)(x^2-19)$. Now can this be reduced further? Well,

$\displaystyle x^2-19=(x-\sqrt{19})(x+\sqrt{19})$

but,

$\displaystyle x^2+4$ cannot be. - Jul 13th 2006, 09:40 AMThePerfectHackerQuote:

Originally Posted by**harold**

What you can do, is reducible (factor) the polynomials in $\displaystyle \mathbb{C}$. Then since, $\displaystyle \mathbb{R} \leq \mathbb{C}$ 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. - Jul 13th 2006, 06:00 PMBubbleBrain_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. - Jul 13th 2006, 07:04 PMThePerfectHackerQuote:

Originally Posted by**harold**

Thus, you end up with a quadradic,

$\displaystyle y^2-15y-75$ - Jul 15th 2006, 11:12 PMmalaygoelQuote:

Originally Posted by**harold**

$\displaystyle x^4 - 15x^2 - 75$

$\displaystyle =(x^2 - \frac{15 + \sqrt{525}}{2})(x^2 - \frac{15 - \sqrt{525}}{2})$

$\displaystyle =(x^2 - \frac{15 + \sqrt{525}}{2})(x^2 + \frac{\sqrt{525} - 15}{2})$

$\displaystyle =(x-\sqrt{\frac{15 + \sqrt{525}}{2}})$$\displaystyle (x+\sqrt{\frac{15 + \sqrt{525}}{2}})$$\displaystyle (x-i\sqrt{\frac{\sqrt{525}-15}{2}})$$\displaystyle (x+i\sqrt{\frac{\sqrt{525}-15}{2}})$

Keep Smiling

Malay - Jul 17th 2006, 08:04 PMmalaygoelQuote:

Originally Posted by**harold**

Quote:

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

$\displaystyle (x-\sqrt{\frac{15 + \sqrt{525}}{2}})$$\displaystyle (x+\sqrt{\frac{15 + \sqrt{525}}{2}})$$\displaystyle (x-i\sqrt{\frac{\sqrt{525}-15}{2}})$$\displaystyle (x+i\sqrt{\frac{\sqrt{525}-15}{2}})$

is written as a product of irreducibles in $\displaystyle \mathbb C[x]$. (So its irreducible over $\displaystyle \mathbb C[x]$)

Quote:

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

Quote:

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

Please someone check it. I am not very confident.

Keep Smiling

Malay