 ## 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 \$\$<a+ib>={(x+iy)(a+ib): x,y in Z}.\$\$

We now consider the quotient ring \$\$Z[i]/<a+ib>={(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 <a+ib>\$\$ 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 <a+ib>.\$\$ 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 <d(p^{2}+q^{2})>.\$\$

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

Hence, \$\$nin<N>.\$\$

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

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

Result B: –      For any \$\$nin Z\$\$ , \$\$niin <a+ib>\$\$ if and only if \$\$nin <N>.\$\$

Proof: –           Similar to above.

Observation: –

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

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

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

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

The following result is further more generalized:

Result C:-  \$\$Z[i]/<a+ib>={r+is+<a+ib> : r,sin Z ~~and~~ 0le r<N ~~and~~ 0le s<d}.\$\$

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

We take any element \$\$x+iy+<a+ib>\$\$ of \$\$Z[i]/<a+ib>.\$\$ By division algorithm, we can write \$\$y=qd+s\$\$ where \$\$q,sin Z, 0le s<d.\$\$ 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+<a+ib>=(x+qbu-qav)+is+iq(a+ib)(u-iv)+<a+ib>\$\$

\$\$=(Nw+r)+is+<a+ib>\$\$ , where \$\$0le r<N\$\$ ,   (division algorithm)

\$\$=r+is+<a+ib>\$\$ where \$\$0le r<N\$\$ , \$\$0le s<d\$\$ (since \$\$Nwin <a+ib>.\$\$)

Hence the result

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

Proof: –           Let \$\$A={r+is+<a+ib>:r,sin Z and 0le r<N and 0le s<d}\$\$

Let \$\$x_{1}+iy_{1}+<a+ib>\$\$ and \$\$x_{2}+iy_{2}+<a+ib>\$\$ be any two members of \$\$A\$\$.

Since \$\$0le y_{1}, y_{2} <d\$\$ and \$\$0le x_{1}, x_{2}<N\$\$  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 <a+ib>\$\$

i.e.  if \$\$( x_{1}-x_{2})+i(y_{1}-y_{2})in <a+ib>\$\$

\$\$=> (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<a+ib>\$\$ 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]/<a+ib>={r+is+<a+ib> : r,s in Z and 0le r<N and 0le s<d}\$\$

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

3)   If \$\$gcd(a,b)=1\$\$ , then \$\$Z[i]/<a+ib>={x+<a+ib>: 0le x<a^{2}+b^{2}}\$\$

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]/<a+ib>\$\$ is a field and so \$\$<a+ib>\$\$ is maximal.

5)   Another way of representing \$\$Z[i]/<a+ib>\$\$ is

\$\$Z[i]/<a+ib>={ r+is+<a+ib> : r,s in Z and 0le r<d and 0le s< N}\$\$

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]/<a+ib>\$\$

We can try looking for other such smarter ways !!

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