# Thread: Quotient ring and field of 9 elements

1. ## Quotient ring and field of 9 elements

Hi guys, I need some help with the following exercise

Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.

J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\$

I know the quotient is a field but I don´t know why is has 9 elements.

2. Originally Posted by orbit
Hi guys, I need some help with the following exercise

Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.

J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\$

I know the quotient is a field but I don´t know why is has 9 elements.

Hmmm....how exactly do you know the quotient is a field?

Tonio

3. Originally Posted by tonio
Hmmm....how exactly do you know the quotient is a field?

Tonio
Because J es a maximal ideal.

4. Originally Posted by orbit
Hi guys, I need some help with the following exercise

Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.

J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\$

I know the quotient is a field but I don´t know why is has 9 elements.
It's not that hard to literally exhibit the elements of the ring. I mean, if you look at the cosets aren't they just $\left\{a+bi+J:a,b\in\{0,1,2\}\right\}$?

5. sorry, I dont get it.

6. Originally Posted by orbit
sorry, I dont get it.
A quotient ring, if anything, is the quotient group of the abelian group underlying the ring, right? So, in particular by definition $\left|\mathbb{Z}[i]/J\right|=\left(\mathbb{Z}[i]:J\right)$ (the index), but I think you'll find that $\mathbb{Z}[i]/J=\left\{a+bi:a,b\in\{0,1,2\}\right\}$.

7. The part I dont understand is why a,b belongs to {0,1,2}....

8. suppose a = k+3m, b = n+3r.

then doesn't a+bi + J = k+ni + J?

9. Originally Posted by Deveno
suppose a = k+3m, b = n+3r.

then doesn't a+bi + J = k+ni + J?
Yes, are the same... But... what do I know from that fact?

10. what you know is that the coset a+bi + J is the same as a(mod 3) + b(mod 3)i + J. that is a,b = 0,1 or 2 (mod 3). this means we get exactly 9 (distinct) elements:

0+0i + J, 0+i + J, 1+0i + J, 1+i + J, 0+2i + J, 2+0i + J, 1+2i + J, 2+i + J, 2+2i + J.

11. Originally Posted by Deveno
what you know is that the coset a+bi + J is the same as a(mod 3) + b(mod 3)i + J. that is a,b = 0,1 or 2 (mod 3). this means we get exactly 9 (distinct) elements:

0+0i + J, 0+i + J, 1+0i + J, 1+i + J, 0+2i + J, 2+0i + J, 1+2i + J, 2+i + J, 2+2i + J.
Ok, thank u so much!

12. Originally Posted by orbit
Ok, thank u so much!

I sitll can´t figure out how did you manage to prove that ideal is maximal without finding out a little more

about how its elements look and, thus, how easy it is to deduce that we have a field with 9 elements...

Tonio

13. as Z[i] is an integral domain, to show J is maximal it suffices to show that J is:____?