## Inverses in algebraic number fields

There is one thing that you haven’t discussed in detail in the lectures is the computation of the inverse of any given nonzero element alpha in an algebraic number field. In question 2 of Exercise Sheet 2, where the number field is $k = \mathbb{Q}[X]/(x^3-2),$ isomorphic to $F = \mathbb{Q}(2^{1/3}),$ I managed to work out that the inverse of the element $X + (f) \in k$, is $(1/2)x^2 + (f),$ by calculating in $F,$ where the basis is $\{1, 2^{1/3}, 2^{2/3}\}.$ But for the other one, $x + x^2 + (f),$ I could not find it. In this field $k$, which we can think of also as $F,$ we know any element beta is some linear combination $a + b(2^{1/3}) + c(2^{2/3})$ of the basis elements, for unique rationals $a,b,c.$ The element $x + x^2 + (f) \in k$ uniquely corresponds to the element $2^{1/3} + 2^{2/3} \in F,$ and to find the inverse of $x + x^2 + (f)$ we need to find rationals $a,b,c$ such that the inverse $(2^{1/3}+2^{2/3})^{-1},$ which is also an element of $F,$ is the linear combination $a +b2^{1/3}+ c2^{2/3}.$ I stopped there, but could not continue further.

The same kind of question is asked in question 3. Is there a general method for computing the inverse of a given element in algebraic number fields $Q[x]/(f)$?

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The general method uses Bezout’s lemma. Given two relatively prime polynomials $f$ and $g$, there are polynomials $a$ and $b$ such that

$af+bg=1$

Furthermore, $a$ and $b$ can be computed explicitly by using Euclid’s algorithm for polynomials. Once you know how to do this, you can take a non-zero class $[g]\in Q[x]/(f)$. Since $f$ is irreducible, $f$ and $g$ must be relatively prime. Then computing $a$ and $b$ as above will give the equation

$[b][g]=1$

in $Q[x]/(f)$.

Try the problems out again from this point of view, and ask again if you get stuck. The ease of doing this kind of polynomial algebra is one of the main reasons that the ‘quotient of polynomial ring’ approach is often more useful than the ‘subfield of $\mathbb{C}$‘ approach to algebraic number fields.