# Thread: Urgently need help! If any1 can do this plzzz help! ><

1. ## Urgently need help! If any1 can do this plzzz help! ><

Let Zp[i] denote the ring {a + bi | a,b E [Z]p, i^2=-1}.

a) Show that if p is not prime, then Zp[i] is not an integral domain.
b) Suppose p is prime. Show that every non-zero element of Zp[i] is a unit iff x^2 + y^2 does not equal 0modp for any pairs x,y E [Z]p.

(Its a baby 'm' and 'p' next to the z's... couldn't find how to shrink letters! :S)

2. Originally Posted by carmz
Let Zp[i] denote the ring {a + bi | a,b E [Z]p, i^2=-1}.

a) Show that if p is not prime, then Zp[i] is not an integral domain.
b) Suppose p is prime. Show that every non-zero element of Zp[i] is a unit iff x^2 + y^2 does not equal 0modp for any pairs x,y E [Z]p.

(Its a baby 'm' and 'p' next to the z's... couldn't find how to shrink letters! :S)
i really don't know if this is correct, however, this is my idea (at least for part a).

if p is not prime, then at least $\mathbb{Z}_p$ is not an integral domain since $p=mn$ where for some m,n both not units (or not equal to 1) and hence $\mathbb{Z}_p$ have zero divisors. Note that $\mathbb{Z}_p \subset \mathbb{Z}_p[i]$ by taking $b=0$, thus $\mathbb{Z}_p[i]$ cannot be an integral domain.

3. Am I right in thinking that the notation [Z]n for example means that if x=n, then x=0?

4. If so, for question 1:

Let p=mn.
Then m,n are in [Z]p.
Hence (m+mi), (n+ni) are in Zp[i]
(m+mi)(n+ni)
=mn+mni+mni-mn
=mni
=0 (since mn=0)

Hence there are zero divisors and it is not an integral domain.

5. For question 2.

Suppose that every non-zero element is a unit (assumtion 1).
Assume there are x,y in [Z]p such that x^2+y^2=0modp and get a contradiction:
(x+yi)(x-yi)=x^2+y^2=0
But these are in Zp[i] and so there exist z1,z2 such that z1(x+yi)=1 and z2(x-yi)=1 by assumption 1.
Hence:
1=1.1=z1(x+yi)z2(x-yi)=z1z2(x+yi)(x-yi)=z1z2(x^2+y^2)=z1z20=0
Therefore every non-zero element a unit implies there are no s,y such that x^2+y^2=0modp

6. Now suppose x^2+y^2 does not equal zero for any x,y pairs.

Show that if x,y are in [Z]p then:
x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p
(can't remember how you do this, but it is true and must be quite straightforward)
Therefore:
(x-yi)/(x^2+y^2) is in Zp[i]

Then for all x+yi we can multiply by (x-yi)/(x^2+y^2) to get:
(x+yi)(x-yi)/(x^2+y^2)=1
So x+yi is a unit.

So x^2+y^2 not equal to zero implies all x+yi are units.

7. "Show that if x,y are in [Z]p then:
x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p
(can't remember how you do this, but it is true and must be quite straightforward)"

You say:
x,y in Z[p]
implies:
x^2 and y^2 in Z[p]
implies
x^2+y^2 in Z[p]
implies
1/(x^2+y^2) in Z[p] (since Z[p] is an integral domain, the tricky step!)
implies
x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p