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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.

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How do you define $\frac{a}{b}$? You could define it as $a\bar b \cdot (b\bar b)^{-1}$, and then it exists, but if you use a different definition you calculation only makes sense if the left side exists.. So in order to use this to prove the existence, you need to assume the existence.... –  N. S. Jan 18 '12 at 17:47
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This seems to make a bit of a meal out of the fact that the inverse of $a+bi$ is $\frac{a-bi}{a^2+b^2}$: the road to existence seems pretty short! –  Pete L. Clark Jan 18 '12 at 20:20

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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.

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The key point is that rationalizing (here real-izing) denominators allows one to lift "existence of inverses of elements $\ne 0\,$" from $\mathbb R$ to $\mathbb C.\:$ Namely, since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\ \Rightarrow\ r^{-1}\in \mathbb R,\:$ so

$$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne\alpha\alpha' = r\in \mathbb R\ \ \Rightarrow\ \frac{1}\alpha\, =\, \frac{\alpha'}{\alpha\:\alpha'}\, =\, \frac{\alpha'}r\in\mathbb C $$

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

More generally one may construct "rational" ($\in$ Z) multiples $\ne 0$ of algebraic elements $\alpha\ne 0$ (of a domain D algebraic over a subring Z) via the constant term of a minimal polynomial of $\alpha$ over Z (vs. above use of norm = product of conjugates). Namely, since $\rm\:0\ne\alpha\in$ D is algebraic over 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 of a domain are cancellable. Thus we may write $\rm\:f(x) = x\ g(x)-n\:$ 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" ($\in$ Z) multiple $\ne 0\,$ of $\rm\,\alpha.\,$ As above, this enables us to "rationalize" a denominator $\rm\:\alpha\in D\:$ by multiplying by $\rm \alpha' = g(\alpha), $ viz.

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

In particular, if domain D is an algebraic extension of a field F, then D is a field, since, as shown, every element $\ne 0\,$ of D divides an element $\ne 0\,$ of F. But elements $\ne 0\,$ of the field F are units (which remain units in 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.

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