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We define the set:

$$\mathbb{Q}[\sqrt[3]2]=\{a_{0}+a_{1}\sqrt[3]{2}+a_{2}\sqrt[3]{2^{2}}:a_{0}, a_1,a_2\in\mathbb{Q}\}$$

It's easy to prove all the properties of fields, except for the unit elements.

So, how can we prove that $$\forall x\in\mathbb{Q^{*}}[\sqrt[3]2],\exists x^{-1} \in\mathbb{Q}[\sqrt[3]2]:xx^{-1}=1$$

And how can we prove this in general for the set $$\mathbb{Q}[\sqrt[n]2]=\{a_{0}+a_{1}\sqrt[n]{2}+a_{2}\sqrt[n]{2^{2}}+...+a_{n-1}\sqrt[n]{2^{n-1}}:a_{0}, a_1,a_2,...,a_{n-1}\in\mathbb{Q}\}$$

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  • $\begingroup$ See also math.stackexchange.com/a/1277313/589. $\endgroup$ – lhf Mar 29 '16 at 1:33
  • $\begingroup$ @Ihf the first link doesn't address the second question or provide much scope for generalisation. (The second one is useful!) $\endgroup$ – Mathmo123 Mar 29 '16 at 1:40
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It's quite fiddly to do this directly, especially in general, although it is possible. Effectively, you have to come up with a generalisation of the idea of "rationalising the denominator". It turns out that $$\frac{1}{a+b\sqrt[3]2+c\sqrt[3]4}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]2+(b^2-ac)\sqrt[3]4}{a^3+2b^3+4c^3-6abc}.$$ Clearly, this isn't something you want to work out from scratch, and I don't fancy trying to generalise this.

As is often the case, it is better is to work abstractly.

First show that $\mathbb Q[\sqrt[3]2]\cong\mathbb Q[X]/(X^3-2)$. To show the latter is a field, all we need to do is show that $(X^3-2)$ is a maximal ideal of $\mathbb Q[X]$. Since $\mathbb Q[X]$ is a PID, this is equivalent to showing that the polynomial $X^3-2$ is irreducible.

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  • $\begingroup$ sorry, I don't follow where you are considering a vector space, and where you are considering a ring. when you say $\mathbb Q(\sqrt[3]2)\cong\mathbb Q[X]/(X^3-2)$, the LHS is a field (by definition) ? and the $\cong$ is in the sense of $\mathbb Q$ vector fields ? $\endgroup$ – reuns Mar 29 '16 at 0:58
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    $\begingroup$ I'm not sure where vector fields come into this. This is an isomorphism of rings. $\endgroup$ – Mathmo123 Mar 29 '16 at 0:59
  • $\begingroup$ the set he defined above $\mathbb{Q}[\sqrt[3]{2}]$ is a $\mathbb Q$ vector space (of dimension $3$), whereas $\mathbb{Q}(\sqrt[3]{2})$ is by definition a field (where we include $\mathbb{Q}$, then $\sqrt[3]{2}$ and then all the needed elements to make it a field) $\endgroup$ – reuns Mar 29 '16 at 1:00
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    $\begingroup$ so whenever $P(X)$ is irreductible in $\mathbb{Q}[X]$, then $\mathbb{Q}[X] / (P(X))$ is the quotient of some PID by some maximal ideal, hence it is a field ? then with $\mathbb Q[\sqrt[3]2]\cong\mathbb Q[X]/(X^3-2)$ (where $\cong$ means rings isomorphism) we get that $\mathbb Q[\sqrt[3]2]$ is a field iff $(X^3-2)$ is a maximal ideal ? $\endgroup$ – reuns Mar 29 '16 at 1:12
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    $\begingroup$ Yes exactly. There are details to check here (why is $(P(X))$ maximal, why does being a PID matter, why are these rings isomorphic) but that is the key idea. $\endgroup$ – Mathmo123 Mar 29 '16 at 1:13
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In the case of the cubic ring $\Bbb Q[\lambda]$ where $\lambda^3=2$, it’s really quite easy to find the reciprocal of $z=a+b\lambda+c\lambda^2$ by the method of rationalizing the denominator. You use the fact that this quantity has the two conjugates $z'=a+b\omega\lambda+c\omega^2\lambda^2$ and $z''=a+b\omega^2\lambda+c\omega\lambda^2$, where $\omega$ is a primitive cube root of unity, $\omega^2+\omega+1=0$.

Just as you expect, you get $$\frac1{z}=\frac{z'z''}{zz'z''}\,,$$ where the numerator is in $\Bbb Q[\lambda]$ and the denominator is in $\Bbb Q$: the resulting formula is exactly the one that @Mathmo123 has given. Since I have done exactly this computation purely by hand many times when I was in graduate school, I do not consider it at all fiddly, but rather pleasingly instructive.

In general, I recommend strongly that you not shrink from hand computation.

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Doing this for $\sqrt[3]{2}$ is a waste of time. ;-) Exactly because doing it for $\sqrt[n]{2}$ would lead to gigantic computations.

Suppose $r\in\mathbb{C}$ is algebraic over $\mathbb{Q}$. We want to see that the set $\mathbb{Q}[r]$ consisting of all the expressions of the form $a_0+a_1r+\dots+a_nr^n$ is a field.

Let $r$ be an algebraic element over the field $F$. Then $F[r]$ is precisely the image of the ring homomorphism $\varphi\colon F[X]\to F[r]$ which is the identity on $F$ and $\varphi(X)=r$.

By the homomorphism theorem, $$ F[r]\cong F[X]/\ker\varphi $$ Now, if $f(X)$ is the minimal polynomial for $r$ over $F$, we can easily see that $\ker\varphi=(f(X))$, the principal ideal generated by $f(X)$.

As $f(X)$ is irreducible, $(f(X))$ is a maximal ideal, so $F[X]/(f(X))$ is a field.


For the particular case, consider the map $\mathbb{Q}[\sqrt[3]{2}]\to\mathbb{Q}[\sqrt[3]{2}]$ given by $t\mapsto t(a+b\sqrt[3]{2}+c\sqrt[3]{4}\,)$. It is a $\mathbb{Q}$-linear map and, with respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, its matrix is $$ \begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{bmatrix} $$ whose inverse (one of the few cases where using the adjugate is simpler than other methods) is $$ \frac{1}{a^3+2b^3+4c^3-6abc} \begin{bmatrix} a^2 - 2bc & -2ac + 2b^2 & -2ab + 4c^2 \\ -ab + 2c^2 & a^2 - 2bc & -2ac + 2b^2 \\ -ac + b^2 & -ab + 2c^2 & a^2 - 2bc \end{bmatrix} $$ which, by the way, proves that $a^3+2b^3+4c^3-6abc\ne0$ as soon as one among $a$, $b$ and $c$ is nonzero.

The inverse of $a+b\sqrt[3]{2}+c\sqrt[3]{4}\ne0$ is thus $$ \frac{(a^2-bc)+(-ab+2c^2)\sqrt[3]{2}+(-ac+b^2)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc} $$

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