A closer look into a well-known quotient ring

The ring $$Z[i]=a+ib:a,binZ,i^2=-1$$  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 $$a+ibinZ[i]$$ and so can be taken in the form $$<a+ib>=(x+iy)(a+ib):x,yinZ.$$

We now consider the quotient ring $$Z[i]/<a+ib>=(x+iy)+(a+ib):x,yinZ$$  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 $$x+iyinZ[i].$$

$$=>n=d(p+iq)(x+iy)$$ , 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 $$kinZ.$$

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 $$ninZ$$ , $$niin<a+ib>$$ 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 $$ninZ$$ then by division algorithm, there exist unique $$q,rinZ$$ with $$0ler<N$$ 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 $$ninZ$$ , can be reduced to the form $$r+<a+ib>$$ for a unique integer $$r$$ satisfying $$0ler<N.$$

The following result is further more generalized:

Result C:-  $$Z[i]/<a+ib>=r+is+<a+ib>:r,sinZand0ler<Nand0les<d.$$

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 $$q,sinZ,0les<d.$$ Again since $$d=gcd(a,b)$$ so there exist $$u,vinZ$$ 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 $$0ler<N$$ ,   (division algorithm)

$$=r+is+<a+ib>$$ where $$0ler<N$$ , $$0les<d$$ (since $$Nwin<a+ib>.$$)

Hence the result

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

Proof: –           Let $$A=r+is+<a+ib>:r,sinZand0ler<Nand0les<d$$

Let $$x_1+i{y}_1+<a+ib>$$ and $$x_2+i{y}_2+<a+ib>$$ 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 $$(x_1+i{y}_1)-(x_2+i{y}_2)in<a+ib>$$

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 $$m,ninZ$$

$$=>(x_1-x_2)=am-bn$$ 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)   $$Z[i]/<a+ib>=r+is+<a+ib>:r,sinZand0ler<Nand0les<d$$

2)   $$Z[i]/<a+ib>$$ contains exactly the above $$Ntimesd=a^2+b^2$$ elements.

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

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

$$Z[i]/<a+ib>=r+is+<a+ib>:r,sinZand0ler<dand0les<N$$

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.