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This is an exercise in the textbook "Algebra" by Mark Sepanski.

Let $z=n+mi \in \mathbb Z [i], z\ne 0$ and let $I=(z)$. This exercise shows that $|\mathbb Z [i]/I|=N(z)$.

  1. Let $d=(n,m)$. Show $\frac{N(z)}{d}\in I$.
  2. Choose $x,y\in \mathbb Z$ so that $nx+my=d$. Let $u=(y+ix)z$. Show that $u\in I$ with $\Im (u)=d$.
  3. Show that every coset $\mathbb Z[x]/I$ has a representative of the form $a+ib$ with $a\in [0, N(z)/d)$ and $b\in [0,d)$. Hint: Use part (2) and then (1).
  4. Suppose $(a+ib)+I= (a'+ib')+I$ with $a,a'\in [0, N(z)/d)$ and $b,b'\in [0,d)$. Show $b=b'$.
  5. Show that $I\cap\mathbb R= (N (z)/d)\mathbb Z$. Conclude that $a=a'$ and so each representative is unique.

I was able to prove parts 1,2,4. But I was unable to prove part (3) and $I\cap\mathbb R= (N (z)/d)\mathbb Z$.

EDIT: Here's my attempt of part $3$.

From part $2$, we can write $u=p+id$ where $p\in \mathbb Z$.

Let $b+ia \in \mathbb Z$. Then by division algorithm there exists integers $q_1, r_1$ satisfying $a=q_1d+r_1$.

Now, we can write $b+ ia = (b-q_1p)+ q_1 u + r_1i$.

By division algorithm, there exists integers $q_2 , r_2$ satisfying $b-q_1 p = q_2 N(z)/d +r_2$ where $r_2 \in [0, N(z)/d)$.

So we have $b+ia=(q_2 N(z)/d + q_1 u) + r_2 + r_1 i$. By part $(1)$ and $(2)$, we have $b+ia + I= (r_2 + ir_1) + I$

Is part $(5)$ correct? If $z=2+3i$, then $I \cap \mathbb R= 2\mathbb Z + 3 \mathbb Z = \mathbb Z$ which is different from $\mathbb 13 \mathbb Z$.

Am I making a mistake here? Is my proof of part $(3)$ correct?

ashK
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