0

In early algebra lessons, we are taught some algorithms to rewrite a fraction (in $\mathbb{Q}$) with roots in the denominator as another fraction with no roots (i.e. an integer) in the denominator. This procedure is called "rationalization". For instance:

  • a fraction of the form $\frac{1}{\sqrt{a}}$ can be rationalized as follows: $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{a}{\sqrt{a}} $$
  • a fraction of the form $\frac{1}{a + \sqrt{b}}$ can be rationalized as: $$\frac{1}{a + \sqrt{b}} = \frac{1}{a + \sqrt{b}} \cdot \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}$$
  • etc.

I am looking for an algorithm for the following generalized version of this problem.

Let $E/F$ be a field extension, and let $\alpha \in E$ be algebraic over $F$. Consider the field $F(\alpha)$. Let $n$ be the algebraic degree of $\alpha$ over $F$; then we know that an $F$-basis of $F(\alpha)$ is $\{1, \alpha, \ldots, \alpha^{n - 1}\}$. Given an element $x \in F(\alpha)$ (which can be written as $$x = \sum_{i = 0}^{n - 1} x_i\alpha^i$$ for some unique $x_i \in F$), how can we find the coefficients $y_i \in F$ such that $$y := \sum_{i = 0}^{n-1} y_i\alpha^i \in F(\alpha) $$ is the multiplicative inverse of $x$ (i.e. $x^{-1} = y$)?

(What I mean by algorithm here is a constructive recipe to find the coefficients of the inverse, assuming that we know how to compute the minimal polynomial of $\alpha$ over $F$.)

Rationalization (with a single root) is indeed a special case of this problem: it's the case where $E = \mathbb{R}$, $F = \mathbb{Q}$ and $\alpha$ is some irrational root of an integer. For instance, the second example above consists of $\alpha = \sqrt{b}$ (which has algebraic degree 2 over $\mathbb{Q}$) with $x = a + \sqrt{b}$, and finds $y = \frac{a}{a^2 - b} - \frac{1}{a^2 - b}\sqrt{b}$.

In this question, the accepted answer proves that $\mathrm{span}_F \{1, \alpha, \ldots, \alpha^{n-1}\}$ is indeed a field, proving the existence of multiplicative inverses, but it doesn't explicitly construct the inverse as a linear combination of $\{1, \alpha, \ldots, \alpha^{n-1}\}$. In other words, I'm looking for a constructive proof of the existence of multiplicative inverses in $\mathrm{span}_F \{1, \alpha, \ldots, \alpha^{n-1}\}$.

P.S.

An even more general problem would be the following:

Let $E/F$ be a field extension, and let $\alpha_1, \ldots, \alpha_m \in E$ be algebraic over $F$ and linearly independent in the $F$-vector space $E$. Consider the field $F(\alpha_1, \ldots, \alpha_m)$. Let $n_j$ be the algebraic degree of $\alpha_j$ over $F$ for every $j$; then we know that an $F$-basis of $F(\alpha_1, \ldots, \alpha_m)$ is: $$ \mathcal{B} := \left\{ \prod_{j = 1}^m \alpha_j^{k_j} \mid \forall j \quad 0 \le k_j < n_j \right\} $$ Given an element $x \in F(\alpha_1, \ldots, \alpha_m)$ with coorinates $[x]_\mathcal{B} = (x_{0, \ldots, 0}, \ldots, x_{n_1, \ldots, n_m})$ how can we find the coordinates of $x^{-1}$ (which is again in $F(\alpha_1, \ldots, \alpha_m)$) with respect to the basis $\mathcal{B}$?

But I guess an algorithm for this can be obtained from an algorithm of the case with $m = 1$ considering that $F(\alpha, \beta) = \left(F(\alpha)\right)(\beta)$ (see here).

Anakhand
  • 2,572
  • 2
    $1=xy=\sum x_iy_j\alpha^{i+j}$. Now you can express each power of $\alpha$ as a polynomial of degree at most $n-1$ in $\alpha$, so you get $n$ linear equations for the $n$ unknowns $y_j$. – Gerry Myerson Feb 15 '22 at 11:46
  • 2
    We can efficiently compute inverses of $,f(\alpha)\in F(\alpha)\cong F[x]/p(x),$ (for prime $,p(x)),$ by using the extended Euclidean algorithm in $F[x]$ to compute a Bezout equation $,g(x)f(x) f(x)+ h(x)p(x) = 1,,$ so $,g(\alpha)f(\alpha) = 1,$ in $F(\alpha),,$ just like for integers (or any Euclidean domain), e.g. see here and its links. For other algorithms see Henri Cohen's book: A course in computational algebraic number theory. – Bill Dubuque Feb 16 '22 at 08:56
  • @Bill, you meant to write $g(x)f(x)+h(x)p(x)=1$, right? – Gerry Myerson Feb 16 '22 at 12:20
  • 1
    @Gerry Yes, of course (editing slipup). – Bill Dubuque Feb 16 '22 at 14:22

1 Answers1

0

The following approach is perhaps not the most "algorithmic" for higher degrees (due to the difficulty of finding roots exactly), but I see it as the most direct generalization of rationalization.

Assuming we have an algorithm for computing the minimal polynomial $f$ of $x$ over $F$ (or in fact any polynomial with coefficients in $F$ having $x$ as a root), and assuming we can find all the roots $x_i$ of $f$, then the product of all its roots (including $x_1 = x$) will be the free coefficient (possibly with a negative sign) of $f$, which is in $F$. That is, we have $f(t) = c + tg = \prod_{i = 1}^n (t - x_i)$ so $F \ni c = (-1)^n\prod_i x_i$.

Therefore, in analogy with "high school rationalization", we can write: $$ \frac{1}{x} = \frac{1}{x} \cdot \frac{\prod_{i > 1} x_i}{\prod_{i > 1} x_i} = (-1)^n \frac{\prod_{i > 1} x_i}{c}$$

Anakhand
  • 2,572