# gaussian integers/integral domains

• Nov 27th 2007, 04:28 PM
eigenvector11
gaussian integers/integral domains
I am having some trouble with these problems:

1. Find all solutions of (x^2) - x + 2 = 0 over Z_3[i], where Z_3[i] =
{a + bi, where a, b are an element of Z_3} = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}, where i^2 = -1 (the ring of Gaussian integers modulo 3).

2. Find all units and zero divisors of Z_3 "direct sum" Z_6.

3. Let F be a finite field with n elements. Prove that x^(n-1) = 1 for all nonzero x in F.

4. Is Z_2[i] a field? Is it an integral domain? Note that Z_2[i] = {a + bi, where a, b are an element of Z_2} = {0, 1, i, 1 + i}, where i^2 = -1 (the ring of gaussian integers modulo 2).

For #4, I'm fairly certain it's not an integral domain since (1 + i)(1 + i) = 2i = 0 in mod 2, making (1+i) a zero divisor. I'm not sure how to show if it is a field or not though.

Thanks guys.
• Nov 27th 2007, 07:03 PM
ThePerfectHacker
Quote:

Originally Posted by eigenvector11
1. Find all solutions of (x^2) - x + 2 = 0 over Z_3[i], where Z_3[i] =
{a + bi, where a, b are an element of Z_3} = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}, where i^2 = -1 (the ring of Gaussian integers modulo 3).

Any element can be written as $\displaystyle a+bi$ where $\displaystyle a,b$ are the equivalence classes mod 3. There are 9 such possibilities just check each one.
Quote:

2. Find all units and zero divisors of Z_3 "direct sum" Z_6.
A unit $\displaystyle (a,b)$ in this ring is such that $\displaystyle (a,b)(x,y)=(1,1)\implies (ax,by)=(1,1)$. Thus the pair is invertible if and only if $\displaystyle a \mbox{ and } b$ are invertible. Use this to complete the problem.

For zero divisors note that if $\displaystyle (a,b)(c,d)=(0,0)\implies (ac,bd)=(0,0)$ but $\displaystyle a,c\in \mathbb{Z}_3$ are in a field and they are no zero divisors hence $\displaystyle a,c=0$. But $\displaystyle bd=0$ are in $\displaystyle \mathbb{Z}_6$ so $\displaystyle b=2,d=3 \mbox{ and }b=3,d=2$ are zero divisors.

Quote:

3. Let F be a finite field with n elements. Prove that x^(n-1) = 1 for all nonzero x in F.
The multiplicative set of F of non-zero elements forms a group under multiplication. Since it has n-1 elements it means x^(n-1) = 1 (property of groups).

Quote:

4. Is Z_2[i] a field? Is it an integral domain? Note that Z_2[i] = {a + bi, where a, b are an element of Z_2} = {0, 1, i, 1 + i}, where i^2 = -1 (the ring of gaussian integers modulo 2).
Just check the defitions.