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 data-recalc-dims=={(x+iy)(a+ib): x,y in Z}." />

We now consider the quotient ring Z[i]/<a+ib data-recalc-dims=={(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 data-recalc-dims=" /> if and only if nin <(a^{2}+b^{2}/gcd(a,b)) data-recalc-dims=." />

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

Now let  nin <a+ib data-recalc-dims=." /> This implies that n=(a+ib)(x+iy) for some x+iyin Z[i].

= data-recalc-dims= n=d(p+iq)(x+iy)" /> , which gives py=-qx = data-recalc-dims= 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 = data-recalc-dims= x/p=-y/q=k" /> say, where kin Z.

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

= data-recalc-dims=n=d(p^{2}+q^{2})(k+i0)" /> which shows that nin <d(p^{2}+q^{2}) data-recalc-dims=." />

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

Hence, nin<N data-recalc-dims=." />

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

Therefore nin <a+ib data-recalc-dims=" /> if and only if  nin <(a^{2}+b^{2}/gcd(a,b)) data-recalc-dims=." />

Result B: -      For any nin Z , niin <a+ib data-recalc-dims=" /> if and only if nin <N data-recalc-dims=." />

Proof: -           Similar to above.

Observation: -

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

2)  Since a^{2}+b^{2}=(a+ib)(a-ib) so it is to be noted that <N data-recalc-dims= subseteq ." />

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 data-recalc-dims=" /> so +<a+ib data-recalc-dims==r+." />

Thus any element of the form n+<a+ib data-recalc-dims=" /> , where nin Z , can be reduced to the form r+<a+ib data-recalc-dims=" /> for a unique integer r satisfying 0le r<N.

The following result is further more generalized:


Result C:-  Z[i]/<a+ib data-recalc-dims=={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+<a+ib data-recalc-dims=" /> of Z[i]/<a+ib data-recalc-dims=." /> 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 data-recalc-dims==(x+qbu-qav)+is+iq(a+ib)(u-iv)+" />

=(Nw+r)+is+<a+ib data-recalc-dims=" /> , where 0le r<N ,   (division algorithm)

=r+is+<a+ib data-recalc-dims=" /> where 0le r<N , 0le s<d (since Nwin <a+ib data-recalc-dims=." />)

Hence the result


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

Proof: -           Let A={r+is+<a+ib data-recalc-dims=:r,sin Z and 0le r

Let x_{1}+iy_{1}+<a+ib data-recalc-dims=" /> and x_{2}+iy_{2}+<a+ib data-recalc-dims=" /> 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 data-recalc-dims=" />

i.e.  if ( x_{1}-x_{2})+i(y_{1}-y_{2})in <a+ib data-recalc-dims=" />

= data-recalc-dims= (x_{1}-x_{2})+i(y_{1}-y_{2})=(a+ib)(m+in)" /> for some m,nin Z

= data-recalc-dims=( x_{1}-x_{2})=am-bn" /> and ( y_{1}-y_{2})=bm+an

= data-recalc-dims=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 data-recalc-dims=" /> 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 data-recalc-dims=={r+is+ : r,s in Z and 0le r

2)    Z[i]/<a+ib data-recalc-dims=" /> contains exactly the above Ntimes d=a^{2}+b^{2} elements.

3)   If gcd(a,b)=1 , then Z[i]/<a+ib data-recalc-dims=={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]/<a+ib data-recalc-dims=" /> is a field and so <a+ib data-recalc-dims=" /> is maximal.

5)   Another way of representing Z[i]/<a+ib data-recalc-dims=" /> is

Z[i]/<a+ib data-recalc-dims=={ 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]/<a+ib data-recalc-dims=" />

We can try looking for other such smarter ways !!


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


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