Division of Complex Numbers (by rationalizing the denominator)

Ahlfors says that once the existence of the quotient $\frac{a}{b}$ has been proven, its value can be found by calculating $\frac{a}{b} \cdot \frac{\bar b}{\bar b}$. Why doesn't this manipulation show the existence of the quotient?

$\frac{a}{b} = \frac{a}{b} \cdot \frac{\bar b}{\bar b} = a\bar b \cdot b^{-1} \cdot\bar b^{-1} = a\bar b \cdot (b\bar b)^{-1}$, the last term clearly exists since the thing being inversed is real.

Solution 1:

There's a small logical hitch. If you don't know that $\frac{a}{b}$ exists, then you can't begin algebraically manipulating it as you've down to arrive at your last expression, because you don't know you won't arrive at nonsense.

Logically, this boils down to the fact that $$ X \ \hbox{ implies } Y $$ is a true statement if both X and Y are false. Here you're starting with a statement X: "$\frac{a}{b}$ exists" and concluding that Y: "$\frac{a}{b} = a \bar{b} . (b\bar{b})^{-1}$

So there's still the job of showing $\frac{a}{b}$ does exist.

Solution 2:

Key Idea: multiplying by a conjugate allows us to rationalize (here $\color{#c00}{\textit{real-ize}}$) denominators, lifting "existence of inverses of $\,r\ne 0\,$" from $\mathbb R$ to $\mathbb C,\,$ i.e. since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\, \Rightarrow\, r^{-1}\in \mathbb R,\,$ so

$$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne{\alpha\bar\alpha} = r\in \mathbb R\ \ \Rightarrow \underbrace{\frac{1}\alpha\, =\, \frac{1\ \bar \alpha}{\alpha\,\bar\alpha}\, =\, \frac{\bar\alpha}r\in\mathbb C}_{\textstyle{\color{#c00}{\textit{real-ize}}\ the\ denominator}}\qquad$$

Thus $ $ field $\mathbb R\, \Rightarrow\, $ field $\mathbb C\ $ by using the norm $\rm\,\alpha\to\alpha\!\ \bar\alpha\,$ to lift existence of inverses from $\mathbb R$ to $\mathbb C.$

Generally we can construct "rational" $(\rm \in Z)$ multiples $\ne 0$ of algebraic elements $\,\alpha\ne 0\,$ (of a domain $\rm D$ algebraic over a subring $\rm Z)$ via the constant term of a minimal polynomial of $\alpha$ over $\rm Z$ (vs. above use of norm = product of conjugates). Namely, since $\rm\,0\ne\alpha\in D\,$ is algebraic over $\rm Z,\,$ it is a root of a polynomial $\rm\,0\ne f(x)\in Z[x].\,$ W.l.o.g. $\rm\,f(0)\ne 0\,$ by $\rm\,f(\alpha)\ \alpha^n = 0\ \Rightarrow\ f(\alpha) = 0,\,$ since nonzero elements are cancellable in a domain. Hence $\rm\,f(x) = x\ g(x)-n\,$ for $\rm\ 0 \ne n\in Z.\,$ So $\rm\ f(\alpha) = 0\ \Rightarrow\ \alpha\ g(\alpha) = n\in Z.\,$ So $\rm\,n\,$ is our "rational" $\rm(\in Z)$ multiple $\ne 0\,$ of $\rm\,\alpha.\,$ As above, this enables us to "rationalize" a denominator $\rm\,\alpha\in D\,$ by multiplying by $\,\rm \bar\alpha := g(\alpha),\, $ viz.

$$\rm 0\ne\alpha\in D\ \ \Rightarrow\ \ 0\ne\alpha\alpha' = n\in Z\ \ \Rightarrow\ \ \frac{1}\alpha\ =\ \frac{1\ \bar\alpha}{\alpha\,\bar\alpha}\ = \frac{\bar\alpha}n\in D\qquad $$

This is a prototypical instance of the method of simpler multiples.

To compute such a polynomial $\rm f(x)$ having $\alpha$ as root generally we can use the "determinant trick", i.e. as in Cayley-Hamilton, we compute the characteristic polynomial of the linear map $\rm \,x\mapsto \alpha x\,$ (but there are more efficient algorithms in various contexts).

In particular, if domain $\rm D$ is an algebraic extension of a field $\rm F$, then $\rm D$ is a field, since, as shown, every element $\ne 0\,$ of $\rm D$ divides an element $\ne 0\,$ of $\rm F$. But elements $\ne 0\,$ of the field $\rm F$ are units (which remain units in $\rm D)$, and divisors of units are units. Yours is the special $\rm\ D = \mathbb C,\ \ F = \mathbb R.$

Generalizing the above from domains to rings allows one to conclude that integral (or primitive) extensions cannot increase Krull dimension (= max length of prime ideal chains), see here. Recall that a primitive extension is a ring extension $\rm R \subset E$ where every element of E is primitive over R, i.e. every element of E is a root of a polynomial $\rm\,f(x)\in R[x]\,$ that is primitive, i.e. content($\rm f$) = $1,\,$ i.e. the ideal in R generated by the coefficients of $\rm\,f\,$ is $\rm (1) = R.\,$ One easily shows that an element is primitive over R iff it is a root of a polynomial $\rm\,f\in R[x]\,$ have some coefficient being $1$ (or a unit). Thus primitive extensions are generalizations of integral extensions. Like integral elements, primitive elements satisfy the crucial property that they remain primitive modulo a prime P, since they are roots of a polynomial f with some coefficient $= 1,\, $ so f cannot vanish mod P. Due to this, the above-mentioned proof on Krull-dimension still works for primitive extensions. They play a key role in various characterizations of integral extensions, e.g. see papers by David E. Dobbs, e.g. see here.