Does the following Multiplication have nonzero divisors

Question

If we define $(x,y).(u,v) = (xu - yv,xv + yu)$ do we have any non zero divisors for this meaning can we find non zero elements $(x,y)$ and $(u,v)$ such that $(x,y).(u,v) = (0,0)$ i tried to think of an example but i couldn't find any , any suggestions ?

Answer 1

You could do it indirectly by proving that the complex numbers form a field, but you could also do it directly as follows. If $(x,y) \cdot (u,v) = (xu - yv, xv+yu) = (0,0)$, then $xu = yv$ and $xv = -yu$. Now assume both $(x,y)$ and $(u,v)$ are not zero. If $x \neq 0$, multiply the first equation by $x$ and get $$ x^2 u = yxv = -y^2u \implies (x^2+y^2)u = 0 \implies u = 0. $$ Using $xv = -yu = 0$ and $x \neq 0$, we get $v=0$, which means $(u,v) = 0$. You can do a similar computation if $y \neq 0$, ; I leave it up to you.

Hope that helps,

Answer 2

No you can't. Your operation is equivalent to the multiplication of 2 complex numbers $x+iy$ and $u+iv$. Multiplying any two Complex numbers is equivalent to multiplying their norm and adding the arguments. The product of two non-zero norms is a product of two non-zero Real numbers, and cannot be $0$, since the Reals are an integral domain ( a field, actually).

Answer 3

I don't think so.

NOTE: There are no zero divisors in $(\mathbb{C},+,\times)$

Note that when you consider the multiplication in $\mathbb{C}$

$(x+iy)\cdot (u+iv)=(ux-vy)+i(uy+xv)$

If you consider the map $(x+iy)\to(x,y)$ you can see that the operation here and multiplication in $\mathbb{C}$ are equivalent and note that there is no zero divisor in $\mathbb{C}$ as it is a field and hence an integral domain.