Thread: Irreducible Polynomial

Is the given polynomial irreducible:

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

Thanks. MK.

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
We have,



Thus, it has no zeros.
Thus it is irreducible over .

Do the same for

Why does it turn into -1 instead of -2?

Thanks for the earlier response.

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 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).

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

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
We have,

-1=1" alt="1^2+1-1=1" />
-1=1+2-1=2" alt="2^2+2-1=1+2-1=2" />
Thus, it has no zeros.
Thus it is irreducible over .

Do the same for

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:



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 implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that . 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:
in Z3[x].

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



So yes, it should be reducible.

-Dan

Originally Posted by topsquark

I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:



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 implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that . 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:
in Z3[x].

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