Find a non-zero proper ideal in $\mathbb C\times\mathbb C$

Question

Let $S$ be a ring such that $S=\mathbb C\times\mathbb C$, where $\mathbb C$ is the field of complex numbers. Multiplication on $S$ is given by $$(a,b)\cdot(c,d)=(ac-db, ad+bc)$$

The problem asks to find a non-zero proper ideal in $S$.

I was trying to find this ideal by computing for example $(x+yi, z+wi)\cdot(a+bi, a+bi),$ or similar, as I hoped for getting result again of the form $(a+bi, a+bi)$. Unfortunately no computation lead to this result, so I guess that was not a good idea. I think that maybe I should look at this problem from different view, but do not know from which one. Can anyone give me a hint, please?

Answer 1

Note that the construction of $S$ from $\mathbb{C}$ is basically the same as that of $\mathbb{C}$ from $\mathbb{R}$: you are formally adding a square root of $-1$. The difference is of course that $\mathbb{C}$ already has such a root, so $S$ is isomorphic (but not equal!) to the product ring $\mathbb{C}\times \mathbb{C}$.

To find a non-trivial ideal is the same in this case as finding a non-trivial idempotent: an element $x\in S$ such that $x^2=x$. You can then take $I=xS$. Under an isomorphism $S\simeq \mathbb{C}\times \mathbb{C}$, this amounts to choosing $0\times \mathbb{C}$ or $\mathbb{C}\times 0$.

Now it is not hard to see that $x=(\frac{1}{2},\frac{i}{2})$ is idempotent.

Answer 2

Hint: To find a proper, non-zero ideal of $S$, it is sufficient to find a non-zero element $\alpha$ of $S$ that is not invertible in $S$. Indeed, then the ideal generated by $\alpha$ will be non-zero, but it will also be proper, since it cannot contain $1$. Note that in $S$, $1 = (1, 0)$.

Can you find $a, b \in \mathbb{C}$ not both zero such that for all $c, d \in \mathbb{C}$ we will have $(ac - db, ad + bc) \neq (1, 0)$?

Answer 3

Hint $ $ now we've four square roots of $\,-1\,$ via $\,{\rm i}=(i,0),\, {\rm j} = (0,1)\,$ so $\,\rm 0 = j^2\!-i^2 = (j\!-\!i)(j\!+\!i)\,$ so $\,\rm j\!-\!i\,$ is a zero-divisor so nonunit, so $\,\rm (0) \subsetneq (j\!-\!i)\subsetneq (1)$.

Remark $ $ Generally if a polynomial has more roots than its degree then this yields a zero-divisor, e.g. here we adjoined to $\,\Bbb C\,$ a $\rm\color{#c00}{new}$ root $\rm\,\color{#c00}{j\neq \pm i}\,$ of $\rm \,x^2+ 1 = (x-i)(x+i)\,$ so $\rm \,(j-i)(j+i)=0$.