# Irreducible Polynomial

• Oct 21st 2006, 06:01 PM
MKLyon
Irreducible Polynomial
Is the given polynomial irreducible:

(x^2)+x-2 in Z3, Z7?

Thanks. MK.
• Oct 21st 2006, 08:23 PM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
Is the given polynomial irreducible:

(x^2)+x-2 in Z3, Z7?

Thanks. MK.

Whenever you a given a polynomial over a field of degree three or two simply look if it has zero's to determine if it is reducible.

Check $0,1,2\in \mathbb{Z}_3[x]$
We have,
$0^2+0-2=1$
$1^2+1-1=1$
$2^2+2-1=1+2-1=2$
Thus, it has no zeros.
Thus it is irreducible over $\mathbb{Z}_3[x]$.

Do the same for $\mathbb{Z}_7[x]$
• Oct 21st 2006, 08:43 PM
MKLyon
Why does it turn into -1 instead of -2?

Thanks for the earlier response.
• Oct 21st 2006, 08:47 PM
ThePerfectHacker
Quote:

Originally Posted by MKLyon
Why does it turn into -1 instead of -2?

You mean why does it turn into 1 instead of -2.
Because -2 is defined as the additive inverse in the field $\mathbb{Z}_3[x]$ the additive inverse of 2 is 1 because 1+2=0 (we are in the set {0,1,2} definied addition and multiplication modulo 3).
• Oct 22nd 2006, 08:55 AM
MKLyon
By this, I concluded that the equation was reducible in Z7[x] because 3 and 6 produced zeros. Can anyone confirm this?
• Oct 22nd 2006, 09:08 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Whenever you a given a polynomial over a field of degree three or two simply look if it has zero's to determine if it is reducible.

Check $0,1,2\in \mathbb{Z}_3[x]$
We have,
$0^2+0-2=1$
$1^2+1-1=1" alt="1^2+1-1=1" />
$2^2+2-1=1+2-1=2" alt="2^2+2-1=1+2-1=2" />
Thus, it has no zeros.
Thus it is irreducible over $\mathbb{Z}_3[x]$.

Do the same for $\mathbb{Z}_7[x]$

Quote:

Originally Posted by MKLyon
Why does it turn into -1 instead of -2?

Thanks for the earlier response.

I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:
$1^2+1-2=0$
$2^2+2-2=1+2-2=1$

which shows that the polynomial IS reducible in Z3[x].

(Okay, so the red coloring didn't work in the quote! You still get the idea.)

-Dan

PS: That x = 1 produces $x^2+x-2 = 0$ implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that $x^2 + x - 2 = (x - 1)(x + 2)$. Where is the x + 2 factor in the above list? Well, this would imply a root of x = -2 = 1 (mod 3), so really x = 1 again. This says that:
$x^2 + x - 2 = (x - 1)(x + 2) = (x - 1)^2$ in Z3[x].
• Oct 22nd 2006, 09:12 AM
topsquark
Quote:

Originally Posted by MKLyon
By this, I concluded that the equation was reducible in Z7[x] because 3 and 6 produced zeros. Can anyone confirm this?

Using TPH's method I got that
$1^2 + 1 - 2 = 0$
$5^2 + 5 - 2 = 0$

So yes, it should be reducible.

-Dan
• Oct 22nd 2006, 09:55 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:
$1^2+1-2=0$
$2^2+2-2=1+2-2=1$

which shows that the polynomial IS reducible in Z3[x].

(Okay, so the red coloring didn't work in the quote! You still get the idea.)

-Dan

PS: That x = 1 produces $x^2+x-2 = 0$ implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that $x^2 + x - 2 = (x - 1)(x + 2)$. Where is the x + 2 factor in the above list? Well, this would imply a root of x = -2 = 1 (mod 3), so really x = 1 again. This says that:
$x^2 + x - 2 = (x - 1)(x + 2) = (x - 1)^2$ in Z3[x].

I see what I did wrong now,, thank you.