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


=(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.


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