Proving J[i] is a Euclidean ring.

Today we’ll try to realign our intuition with the standard textbook proof of “J[i] is a Euclidean ring”.

J[i] denotes the set of all complex numbers of the form a+bi where a and b are integers. d(a+bi)=a^2+b^2.

Let us take two complex numbers m+in, c+id\in J[i]. Let m=qc+r_1 and n=wd+r_2, where d(r_1)<d(c) and d(r_2)<d(d). Then if there exists a complex number x+iy such that m+in=(x+iy)(c+id)+(r_1+ir_2), we’ll be done. This is because if d(r_1)<d(c)\implies r_1^2<c^2 and d(r_2)<d(d)\implies r_2^2<d^2, then r_1^2+r_2^2<c^2+d^2, which in turn implies d(r_1+ir_2)<d(c+id). This is what we wanted.


According to the conditions stated, we have xc-yd=qc and xd+yc=wd. From these two equations, we need to determine the values of x and y, and prove x+iy\in J[i]. The equations simplify to x(c^2+d^2)=qc^2+wd^2 and y(c^2+d^2)=cd(w-q). Are x and y integers, provided c,d,q and w are? A quick verification should tell you this is not true (try substituting c=3,d=4,q=2 and w=3). Then is there no such element x+iy\in J[i] such that this condition is satisfied? The fact that we’ve failed to find such an element does not imply one cannot exist. Note that our approach had several restrictions like xc-yd=qc, etc.

Let us now go to the book’s proof. The book first proves it for integers. A noteworthy step is ensuring the absolute value of the remainder is \leq\left|\frac{1}{2}n\right|. Then it takes x\overline{x}, which is a positive integer, and uses it in a particular fashion to prove that a=bq+r where d(r)<d(b). Don’t worry if you don’t get what I’m talking about. This proof is present in its entirety on pg.150 in “Topics in Algebra” by Herstein.

How different is this from our earlier method? Earlier we were concerned with quotients and remainders of m and n. The element x+iy and r_1+ir_2 were, for some reason, thought to be of a pre-set nature; r_1 was thought to be x\mod c and r_2 was thought to be d\mod d. Here, we have no such pre-sets. There are two main techniques here, which are analyzed below:

1. |u_1|\leq\left|\frac{1}{2}n\right| and |v_1|\leq\left|\frac{1}{2}n\right|. What does this do? It ensures that d(u_1+iv_1)<d(n). Had we not ensured these bounds, it is possible that u_1^2+v_1^2 would have exceeded n^2. Note that the bounds of u_1 and v_1 cannot be made much sharper; in other words, we can’t ensure that |u_1|<\leq\left|\frac{1}{3}n\right|, etc.

2. The observation that d(a(b+c))=d(a)d(b+c). This is only possible under the current definition of d. If d(ab)=a+b, then this wouldn’t be true.

Let x+iy,m+in and c+id have the same descriptions as given at the beginning of the article. What ultimately is the nature of x+iy? If m(c^2+d^2)=q(c^2+d^2)+u_1 where |u_1|\leq\left|\frac{1}{2}(c^2+d^2)\right|, then x=q. Similarly, if n(c^2+d^2)=r(c^2+d^2)+v_1 where |v_1|\leq\left|\frac{1}{2}(c^2+d^2)\right|, then y=r. This is different from the nature of x+iy that we’d initially thought of. No surprise then that our approach didn’t work.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: