A closer look into a well-known quotient ring

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

Posted at 17:13h in Articles, English, Others, Problems by Manjil Saikia

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 !!