Is the given polynomial irreducible:

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

Thanks. MK.

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- Oct 21st 2006, 05:01 PMMKLyonIrreducible Polynomial
Is the given polynomial irreducible:

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

Thanks. MK. - Oct 21st 2006, 07:23 PMThePerfectHacker
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 $\displaystyle 0,1,2\in \mathbb{Z}_3[x]$

We have,

$\displaystyle 0^2+0-2=1$

$\displaystyle 1^2+1-1=1$

$\displaystyle 2^2+2-1=1+2-1=2$

Thus, it has no zeros.

Thus it is irreducible over $\displaystyle \mathbb{Z}_3[x]$.

Do the same for $\displaystyle \mathbb{Z}_7[x]$ - Oct 21st 2006, 07:43 PMMKLyon
Why does it turn into -1 instead of -2?

Thanks for the earlier response. - Oct 21st 2006, 07:47 PMThePerfectHacker
- Oct 22nd 2006, 07:55 AMMKLyon
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, 08:08 AMtopsquark
I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:

$\displaystyle 1^2+1-2=0$

$\displaystyle 2^2+2-2=1+2-2=1$

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 $\displaystyle x^2+x-2 = 0$ implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that $\displaystyle x^2 + x - 2 = (x - 1)(x + 2)$. 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:

$\displaystyle x^2 + x - 2 = (x - 1)(x + 2) = (x - 1)^2$ in Z3[x]. - Oct 22nd 2006, 08:12 AMtopsquark
- Oct 22nd 2006, 08:55 AMThePerfectHacker