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

The ring is quite a well- known ring in Algebra. Further, being a principal ideal domain, every ideal of is generated by a single element say and so can be taken in the form ={(x+iy)(a+ib): x,y in Z}." />

We now consider the quotient ring ={(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 and are both non zero. The quantities and deserve special attention in the discussion.

**Result A**: - An integer " /> if and only if ." />

** Proof: **- Since , we can write , where

Now let ." /> This implies that for some

n=d(p+iq)(x+iy)" /> , which gives p|qx" /> and (“|” means divides.)

And since , so and which means that and are integers.

Also, x/p=-y/q=k" /> say, where

Thus we get

n=d(p^{2}+q^{2})(k+i0)" /> which shows that ." />

But

Hence, ." />

Conversely if " /> then by using the relation it can be easily shown that ." />

Therefore " /> if and only if ." />

**Result B****: -** For any , " /> if and only if ." />

**Proof****: -** Similar to above.

**Observation****: -**

1) Since do not belong to " /> so it is clear that the elements ,1+, dots , (N-1)+" /> of " /> are distinct.

2) Since so it is to be noted that subseteq ." />

3) If then by division algorithm, there exist unique with such that Since " /> so =r+." />

Thus any element of the form " /> , where , can be reduced to the form " /> for a unique integer satisfying

The following result is further more generalized:

** **

**Result C****:- **={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 " /> of ." /> By division algorithm, we can write where Again since so there exist such that And it is easy to see that which gives

=(x+qbu-qav)+is+iq(a+ib)(u-iv)+" />

" /> , where , (division algorithm)

" /> where , (since ." />)

Hence the result

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

**Proof****: -** Let :r,sin Z and 0le r

Let " /> and " /> be any two members of .

Since and a simple observation reveals that

1) cannot be a non –zero multiple of and

2) cannot be a non-zero multiple of

The above members will be equal if " />

i.e. if " />

(x_{1}-x_{2})+i(y_{1}-y_{2})=(a+ib)(m+in)" /> for some

( x_{1}-x_{2})=am-bn" /> and

d|( x_{1}-x_{2})" /> and since

i.e. is a multiple of d, but by (1) above i.e.

We thus have " /> which in turn implies that is a multiple of but by (2) above, the only possibility is i.e. Thus the elements of are distinct.

** **

**Conclusion****:- **From the above results it is clear that

1) ={r+is+ : r,s in Z and 0le r

2) " /> contains exactly the above elements.

3) If , then ={x+: 0le x

which is isomorphic to the ring of integers modulo

4) Further if is prime then " /> is a field and so " /> is maximal.

5) Another way of representing " /> is

={ 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 or is zero. In this case we can take and

There may be other better ways of representing the elements of " />

We can try looking for other such smarter ways !!

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

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