Quotient Ring of the Gaussian Integers

**chiph588@** · January 29th 2010, 03:04 PM

Given $p \equiv 3 \mod{4}$ is prime in $\mathbb{Z}$ , show $\mathbb{Z}[i]/(p)$ has order $p^2$ .

**aliceinwonderland** · January 29th 2010, 03:31 PM

Originally Posted by chiph588@

Given $p \equiv 3 \mod{4}$ is prime in $\mathbb{Z}$ , show $\mathbb{Z}[i]/(p)$ has order $p^2$ .

An odd prime p has a form either 1(mod 4) or 3(mod 4). We can see that 1(mod 4) is reducible in Z[i] (For example, 5=(1+2i)(1-2i)). An odd prime p is irreducible iff it has the form 3(mod 4).

We know that Z[i] is an integral domain and a PID. We also know that $p \equiv 3 (mod{4})$ is irreducible. Thus <p> is maximal ideal in Z[i]. It follows that Z[i]/<p> is a field having p^2 elements.

**chiph588@** · January 29th 2010, 05:22 PM

Originally Posted by aliceinwonderland

It follows that Z[i]/<p> is a field having p^2 elements.

I don't quite see how that follows.

**aliceinwonderland** · January 29th 2010, 05:32 PM

Originally Posted by chiph588@

I don't quite see how that follows.

An element of Z[i]/<p> has the form a+bi. The number of choices of a is p and the number of choices for b is p. So it has p^2 elements.

For example, let p =3. Then, Z[i]/<3>= {0, 1,2, i, 2i, 1+i, 1+2i, 2+i, 2+2i}. It has 9 elements. You can check other cases and verify them.

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