Inverse of a complex number (in Hamilton's pair representation)

Question

I've just started looking at ring theory and I am having trouble working with invertible elements.

In my notes we have been told that $R^\times$ is the set of all the invertible elements of $R$, and it is called the group of units of $R$. The set $R^\times$ of invertible elements of $R$ is a multiplicative group for the multiplication operation of $R$.

I have to work with the cartesian product $S:= R\times R$ with $R$ a commutative ring. Where $(a,b)(c,d) = (ac-db, ad+bc)$ and I am trying to show $(a,b)∈ S^\times$ iff $(a,-b)∈ S^\times$. I'm guessing we define $S^\times$ the same way as we define $R^\times$? We also know that $S$ has unit $(1,0)$.

Should I proceed by showing $(a,b)(a,-b)=(1,0)=(a,-b)(a,b)$? I had seen online that for invertible elements $uv=1=vu$.

Thank you!

Answer 1

Hint $ $ It's simple viewed as $\,S \cong R[x]/(x^2\!+1) = R[i]\,$ where $\,i := [x] = x + (x^2\!+\!1)R[x]$

$$\begin{align}{\rm where}\ \ \ \ (a,\ b)*(c,\ d) &\,=\, (ac\!-\!bd,\ \ \ \, ad\!+\!bc)\\[.2em] \to\ \ (a\!+\!bi)(c\!+\!di) &\,=\, \ \, ac\!-\!bd\! +\! (ad\!+\!bc)\,i \end{align}\qquad\qquad$$

generalizing the familiar case: $\,\Bbb R[x]/(x^2\!+1) = \Bbb R[i]\cong \Bbb C\ $ when $\ R = \Bbb R$

Then, as in any algebraic extension, existence of inverses in $R$ lifts to $R[i]$ by rationalizing ("real-izing") the denominator, which is a prototypical instance of the method of simpler multiples, i.e. we multiply $\,\alpha\in R[i]\,$ by its conjugate(s) to obtain its (simpler multiple) norm $\,r\in R,\,$ which has the simplification effect of reducing $\color{#c00}{\rm division\ by\ \alpha}\in R[i]$ to simpler $\color{#0a0}{\rm division\ by\ } r\in R,\,$ i.e.

$\displaystyle 0\ne\alpha\in R[i]\ \ \Rightarrow\ \ 0\ne\alpha\bar\alpha = r\in R\ \ \Rightarrow\!\!\! \underbrace{\color{#c00}{\frac{1}\alpha}\, =\, \frac{1\ \bar\alpha}{\alpha\:\bar\alpha}\, =\, \color{#0a0}{\frac{\bar\alpha}r}\in R[i]}_{\textstyle{\color{c00}{\textit{real-ize}}\ {\rm to}\ R \rm\ the\ denominator}}$

In this quadratic case the "simpler multiple" of $\alpha\in R[i]\,$ is simply its norm $\,N(\alpha) = {\alpha\bar\alpha}\in R$ $$\begin{align}\alpha &= a+bi\\ \Rightarrow\ \bar\alpha &= a-bi\end{align} \Rightarrow\ \alpha\bar\alpha = a^2+b^2 = r\in R\qquad$$

Answer 2

It’s not true in general that $(a,b)(a,-b)=1$.

Note that for $R=\mathbb R$, the multiplication on $S$ is the multiplication on $\mathbb C$, so if you’re familiar with $\mathbb C$, you can use your intuitions for that to guide you. In $\mathbb C$, the inverse of the conjugate is the conjugate of the inverse.

The same is true here: If $(a,b)$ is invertible, then there is $(c,d)$ such that $(ac-db,ad+bc)=(1,0)$; and then $(a,-b),(c,-d)=(ac-(-d)(-b),a(-d)+(-b)c)=(ac-db,ad+bc)=(1,0).$