polynomial rings

• Dec 11th 2009, 05:38 PM
tbird
polynomial rings
hey there.

i'm so confused on how to do this problem the way it is supposed to be done.

For what integers $\displaystyle a$ does $\displaystyle x^2 + x + 1$ divide $\displaystyle x^4 + 3x^3 + x^2 + 6x + 10$ in $\displaystyle ( \mathbb{Z} /a \mathbb{Z} )[x]$.

i know how to test for $\displaystyle a\in \mathbb{Z}.$ reduce all the coefficents to fit in $\displaystyle \mathbb{Z} /a \mathbb{Z}$ and use the division algorithm to test whether it is a zero.

but how in the world do i do it for the general case?

id start with doing it with a=1,2,... until i found some sort of pattern with primes or something. but its finals time right now, and i have no time for that kind of exhaustive calculating for a homework prob. i just know there's gotta be a better way.

thanks!
• Dec 11th 2009, 11:57 PM
tonio
Quote:

Originally Posted by tbird
hey there.

i'm so confused on how to do this problem the way it is supposed to be done.

For what integers $\displaystyle a$ does $\displaystyle x^2 + x + 1$ divide $\displaystyle x^4 + 3x^3 + x^2 + 6x + 10$ in $\displaystyle ( \mathbb{Z} /a \mathbb{Z} )[x]$.

i know how to test for $\displaystyle a\in \mathbb{Z}.$ reduce all the coefficents to fit in $\displaystyle \mathbb{Z} /a \mathbb{Z}$ and use the division algorithm to test whether it is a zero.

but how in the world do i do it for the general case?

id start with doing it with a=1,2,... until i found some sort of pattern with primes or something. but its finals time right now, and i have no time for that kind of exhaustive calculating for a homework prob. i just know there's gotta be a better way.

thanks!

Divide $\displaystyle x^4 + 3x^3 + x^2 + 6x + 10$ by $\displaystyle x^2 + x + 1$ . The residue is $\displaystyle 6x+12$ and it MUST be zero, so....what has $\displaystyle a$ to be??

Tonio
• Dec 12th 2009, 10:56 AM
tbird
Quote:

Originally Posted by tonio
Divide $\displaystyle x^4 + 3x^3 + x^2 + 6x + 10$ by $\displaystyle x^2 + x + 1$ . The residue is $\displaystyle 6x+12$ and it MUST be zero, so....what has $\displaystyle a$ to be??

Tonio

a=-2?

that doesn't seem right. i guessed i missed something?
• Dec 12th 2009, 11:15 AM
tonio
Quote:

Originally Posted by tbird
a=-2?

that doesn't seem right. i guessed i missed something?

Your guess right: you're missing a lot. You've to find $\displaystyle a$ so that the division is possible in the RING $\displaystyle \mathbb{Z}\slash a\mathbb{Z}[x]$ of polynomials over THE FINITE ring (or field) $\displaystyle \mathbb{Z}\slash a\mathbb{Z}$...

Tonio
• Dec 12th 2009, 11:56 AM
tbird
Quote:

Originally Posted by tonio
Your guess right: you're missing a lot. You've to find $\displaystyle a$ so that the division is possible in the RING $\displaystyle \mathbb{Z}\slash a\mathbb{Z}[x]$ of polynomials over THE FINITE ring (or field) $\displaystyle \mathbb{Z}\slash a\mathbb{Z}$...

Tonio

alright...so i'm a 100% lost.

my pathetic attempt:

$\displaystyle 6x + 12$ must be 0.

so, im thinking that 'plug a into x, so that once the equation is solved i can then reduce the final value to fit in $\displaystyle \mathbb{Z}\slash a\mathbb{Z}[x]$'

i ran over a bunch in my head, and if what i said above is right, then factors of 12 = a

so, a = 1, 2, 3, 4, 6, and 12

(the math to my argument for a=6 -- a=x so (6)(6) + 12 = 48, but 48 must live in $\displaystyle \mathbb{Z}\slash 6\mathbb{Z}[x]$ which is 0, voila)

?? right track ??
• Dec 12th 2009, 02:15 PM
tonio
Quote:

Originally Posted by tbird
alright...so i'm a 100% lost.

my pathetic attempt:

$\displaystyle 6x + 12$ must be 0.

so, im thinking that 'plug a into x, so that once the equation is solved i can then reduce the final value to fit in $\displaystyle \mathbb{Z}\slash a\mathbb{Z}[x]$'

i ran over a bunch in my head, and if what i said above is right, then factors of 12 = a

so, a = 1, 2, 3, 4, 6, and 12

(the math to my argument for a=6 -- a=x so (6)(6) + 12 = 48, but 48 must live in $\displaystyle \mathbb{Z}\slash 6\mathbb{Z}[x]$ which is 0, voila)

?? right track ??

Yessir...now, if you people are only interested in fields (so that, for example, the polynomial ring over it will have nice properties like no zero-divisors and stuff) then you'll have to choose only the prime factors of 12...

Tonio
• Dec 12th 2009, 02:35 PM
tbird
Quote:

Originally Posted by tonio
Yessir...now, if you people are only interested in fields (so that, for example, the polynomial ring over it will have nice properties like no zero-divisors and stuff) then you'll have to choose only the prime factors of 12...

Tonio

Aaahhh!! I see.

You just put so much together for me by throwing that fact in there about fields. The light bulb just got brighter. Maybe I wasn't so lost...