# [SOLVED] Algebra-How do i show this is irreducible?

• Nov 30th 2008, 01:58 AM
thegarden
[SOLVED] Algebra-How do i show this is irreducible?
My question asks me to find the minimum polynomial of .... i + root5

I get x^4 - 8x^2 + 36 = 0 as the minimum polynomial

How can i show this is irreducible.
I've tried the rational root test but that would take too long.
There must be another shorter way to do it.
Thanks.
• Nov 30th 2008, 05:03 AM
thegarden
• Nov 30th 2008, 05:39 AM
Krizalid
To show irreducibility, observe that $\displaystyle x^{4}-8x^{2}+36=\left( x^{2}-4 \right)^2+20>0.$

And do not bump.
• Nov 30th 2008, 06:02 AM
thegarden
I dont understand lol how can i show irreducibility from that? Thanks.

And where does that > 0 inequality come from?
• Nov 30th 2008, 06:33 AM
thegarden
If i let y=x^2 so that the polynomial becomes y^2-8y+36=0, then take the discriminant which gives a negative discriminant, hence the quadratic is irreducible, would that also imply that the quartic is irreducible?
• Nov 30th 2008, 07:03 AM
Moo
Quote:

Originally Posted by thegarden
If i let y=x^2 so that the polynomial becomes y^2-8y+36=0, then take the discriminant which gives a negative discriminant, hence the quadratic is irreducible, would that also imply that the quartic is irreducible?

If you're all working over the real numbers, then you're correct.

~~~~~~~~~~~~~~~~~~~~~~~~~~

As for $\displaystyle (x^2-4)^2+20$, we know that a square number is always positive.

hence $\displaystyle (x^2-4)^2+20 \geq 0+20 > 0$

therefore it is never equal to 0 (for real values) and thus it is not reducible.
• Nov 30th 2008, 07:16 AM
thegarden
Is this the same for rational numbers?
I want to show it is irreducible in Q
• Nov 30th 2008, 07:17 AM
Moo
Quote:

Originally Posted by thegarden
Is this the same for rational numbers?

rational numbers are real numbers.

the thing that would change is if you're working over complex numbers (in the form a+ib)
• Nov 30th 2008, 07:21 AM
thegarden
Quote:

Originally Posted by Moo
rational numbers are real numbers.

the thing that would change is if you're working over complex numbers (in the form a+ib)

Of course lol cheers
• Dec 1st 2008, 07:40 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
To show irreducibility, observe that $\displaystyle x^{4}-8x^{2}+36=\left( x^{2}-4 \right)^2+20>0.$

And do not bump.

Quote:

Originally Posted by Moo
If you're all working over the real numbers, then you're correct.

~~~~~~~~~~~~~~~~~~~~~~~~~~

As for $\displaystyle (x^2-4)^2+20$, we know that a square number is always positive.

hence $\displaystyle (x^2-4)^2+20 \geq 0+20 > 0$

therefore it is never equal to 0 (for real values) and thus it is not reducible.

Irreduciblity does not mean not having any zeros.
This does not fully complete the proof. (Worried)
• May 25th 2009, 12:08 AM
mr fantastic
Quote:

Originally Posted by thegarden
My question asks me to find the minimum polynomial of .... i + root5

I get x^4 - 8x^2 + 36 = 0 as the minimum polynomial

How can i show this is irreducible.
I've tried the rational root test but that would take too long.
There must be another shorter way to do it.