## 26 Feb A closer look into a well-known quotient ring

The ring $Z[i]={a+ib: a,b in Z, i^{2}=-1}$  is quite a well- known ring in Algebra. Further, being a principal ideal domain, every ideal of $Z[i]$  is generated by a single element say $a+ib in Z[i]$ and so can be taken in the form $={(x+iy)(a+ib): x,y in Z}.$

We now consider the quotient ring $Z[i]/={(x+iy)+(a+ib): x,y in Z}$  and try to look for a precise way of representing the elements in it. First we suppose that $a$ and $b$ are both non zero. The quantities $d=gcd(a,b)$ and $N=(a^{2}+b^{2})/gcd(a,b)$ deserve special attention in the discussion.

Result A: -      An integer $nin $ if and only if $nin <(a^{2}+b^{2}/gcd(a,b))>.$

Proof: -   Since $d=gcd(a,b)$, we can write $a+ib=d(p+iq)$, where $gcd(p,q)=1.$

Now let  $nin .$ This implies that $n=(a+ib)(x+iy)$ for some $x+iyin Z[i].$

$=> n=d(p+iq)(x+iy)$ , which gives $py=-qx$ $=> p|qx$ and $q|py.$ (“|” means divides.)

And since $gcd(p,q)=1$ , so $p|x$ and $q|y$ which means that $x/p$ and $y/q$ are integers.

Also, $py= -qx$ $=> x/p=-y/q=k$ say, where $kin Z.$

Thus we get $n=d(p+iq)(kp-ikq)$

$=>n=d(p^{2}+q^{2})(k+i0)$ which shows that $nin .$

But $d(p^{2}+q^{2})={(dp)^{2}+(dq)^{2}}/d=( a^{2}+b^{2})/gcd(a,b)=N.$

Hence, $nin.$

Conversely if $nin$ then by using the relation $a^{2}+b^{2}=(a+ib)(a-ib)$ it can be easily shown that $nin.$

Therefore $nin $ if and only if  $nin <(a^{2}+b^{2}/gcd(a,b))>.$

Result B: -      For any $nin Z$ , $niin $ if and only if $nin .$

Proof: -           Similar to above.

Observation: -

1)  Since $1,2,dots ,N-1$ do not belong to $$ so it is clear that the elements $0+,1+, dots , (N-1)+$ of $Z[i]/$ are distinct.

2)  Since $a^{2}+b^{2}=(a+ib)(a-ib)$ so it is to be noted that $ subseteq .$

3)  If $nin Z$ then by division algorithm, there exist unique $q,rin Z$ with $0le r such that $=Nq+r.$ Since $n-r=Nqin $ so $+=r+.$

Thus any element of the form $n+$ , where $nin Z$ , can be reduced to the form $r+$ for a unique integer $r$ satisfying $0le r

The following result is further more generalized:

Result C:-  $Z[i]/={r+is+ : r,sin Z ~~and~~ 0le r

Proof: -  The proof is a just a simple application of a well- known property of GCD.

We take any element $x+iy+$ of $Z[i]/.$ By division algorithm, we can write $y=qd+s$ where $q,sin Z, 0le s Again since $d=gcd(a,b)$ so there exist $u,vin Z$ such that $d=au+bv.$  And it is easy to see that $d=(a+ib)(u-iv)-i(bu-av)$ which gives

$x+iy+=(x+qbu-qav)+is+iq(a+ib)(u-iv)+$

$=(Nw+r)+is+$ , where $0le r ,   (division algorithm)

$=r+is+$ where $0le r , $0le s (since $Nwin .$)

Hence the result

Result D: -      The elements in the set in RHS of D are distinct.

Proof: -           Let $A={r+is+:r,sin Z and 0le r

Let $x_{1}+iy_{1}+$ and $x_{2}+iy_{2}+$ be any two members of $A$.

Since $0le y_{1}, y_{2} and $0le x_{1}, x_{2}  a simple observation reveals that

1)   $x_{1}-x_{2}$ cannot be a non –zero multiple of $N$ and

2)   $y_{1}-y_{2}$ cannot be a non-zero multiple of $d.$

The above members will be equal if $( x_{1}+iy_{1})-( x_{2}+iy_{2})in $

i.e.  if $( x_{1}-x_{2})+i(y_{1}-y_{2})in $

$=> (x_{1}-x_{2})+i(y_{1}-y_{2})=(a+ib)(m+in)$ for some $m,nin Z$

$=>( x_{1}-x_{2})=am-bn$ and $( y_{1}-y_{2})=bm+an$

$=>d|( x_{1}-x_{2})$ and $d|( y_{1}-y_{2})$ since $d=gcd(a,b)$

i.e. $y_{1}-y_{2}$ is a multiple of d, but by (1) above $y_{1}-y_{2}=0$ i.e. $y_{1}=y_{2}$

We thus have $x_{1}-x_{2}in$ which in turn implies that $x_{1}-x_{2}$ is a multiple of $N$ but by (2) above, the only possibility is $x_{1}-x_{2}=0$ i.e. $x_{1}=x_{2}.$ Thus the elements of  are distinct.

Conclusion:-   From the above results it is clear that

1)   $Z[i]/={r+is+ : r,s in Z and 0le r

2)   $Z[i]/$ contains exactly the above $Ntimes d=a^{2}+b^{2}$ elements.

3)   If $gcd(a,b)=1$ , then $Z[i]/={x+: 0le x

which is isomorphic to the ring $Z_{a^{2}+b^{2}}$ of integers modulo $a^{2}+b^{2}.$

4)   Further if $a^{2}+b^{2}$ is prime then $Z[i]/$ is a field and so $$ is maximal.

5)   Another way of representing $Z[i]/$ is

$Z[i]/={ r+is+ : r,s in Z and 0le r

6)   It can be proved that the same result holds for the case when one of $a$ or $b$ is zero. In this case we can take $gcd(a,0)=mod(a)$ and $gcd(b,0)=mod(b).$

There may be other better ways of representing the elements of $Z[i]/$

We can try looking for other such smarter ways !!

- Debashish Sharma, Junior Research Fellow, Dept. of Mathematics,NIT Silchar.