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Is $\mathbb{Q}[2^{1/3}]=\{a+b2^{1/3}+c2^{2/3};a,b,c \in \mathbb{Q}\}$ a field?

I have checked that $b2^{1/3}$ and $c2^{2/3}$ both have inverses, $\frac{2^{2/3}}{2b}$ and $\frac{2^{1/3}}{2c}$, respectively.

There are some elements with $a,b,c \neq 0$ that have inverses, as $1+1*2^{1/3}+1*2^{2/3}$, whose inverse is $2^{1/3}-1$.

My problem is that is that I can't seem to find a formula for the inverse, but I also can't seem to find someone who doesn't have an inverse.

Thanks for your time.

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In general, for $\alpha\in\mathbb C$, we have $\mathbb Q[\alpha]$ is a field iff $\alpha$ is an algebraic number, in which case of course $\mathbb Q[\alpha]=\mathbb Q(\alpha)$. –  lhf Dec 21 '11 at 16:21

4 Answers 4

up vote 13 down vote accepted

HINT $\ $ Since $\rm\:f(x) = x^3 - 2\:$ is irreducible in the Euclidean domain $\rm\:\mathbb Q[x]\:,\:$ any lower degree polynomial $\rm\:g(x) = a + b\:x + c\:x^2\:$ is coprime to it in $\rm\:\mathbb Q[x]\:,\:$ therefore, by the Euclidean algorithm, there is a Bezout equation $\rm\ a\ f + b\ g = 1\ $ which yields $\rm\ b(2^{1/3})\ g(2^{1/3}) = 1\ $ for $\rm\:x = 2^{1/3}\:.$

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All answers were helpful, but your hint was very elegant. –  Leonardo Fontoura Aug 21 '11 at 0:35

Actually, the inverse of $x=2^{1/3}$ can be computed in a very elementary way as follows. Since $x\neq0$, the equality $$ x^3-2=x\left(x^2-\frac2x\right)=0 $$ yields $$ \frac1x=\frac12x^2\in{\Bbb Q}[x]. $$ This "trick" extends immediately to any algebraic number.

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Does this differ from the answer I gave? –  Gerry Myerson Aug 21 '11 at 13:02
    
Of course it's the same argument, but I'd say that my choice of wording makes clear (to the non sophisticated student) that the task of computing the inverse of an algebraic number satisfying a given relation does not require the knowledge of the concept of "field" or that of "linear dependence". –  Andrea Mori Aug 21 '11 at 15:02
    
true. But to find the relation satisfied by, say, $2+(3)(2^{1/3})+(4)(2^{2/3})$, even just to know that it does satisfy some relation, that's where those other concepts are going to come in very handy. –  Gerry Myerson Aug 22 '11 at 13:41

A neat way to confirm that it is a field:

Since $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is a PID. $\mathbb{Q}[2^{1/3}] \cong \mathbb{Q}[x] / (x^3 - 2)$. Now, $x^3 - 2$ is irreducible over $\mathbb{Q}$, since if it weren't, there would be a rational root to $x^3 - 2$. Because $\mathbb{Q}[x]$ is a PID and the polynomial is irreducible over $\mathbb{Q}$, $(x^3 - 2)$ is a maximal ideal in $\mathbb{Q}[x]$.

By the Correspondence Theorem of Ideals, we see that as $(x^3 - 2)$ is maximal, $\mathbb{Q}[x] / (x^3 - 2)$ must be a field.

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This might be a stupid question, but isn't $\mathbb{Q}[x]/(x^3 - 2)$ isomorphic to $\mathbb{Q}$ with all the roots of $x^3-2$ adjoined rather than just $\sqrt[3]{2}$? –  kahen Aug 21 '11 at 11:34
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@kahen, no, you just get one root. If you want a fuller discussion, post as a new question. –  Gerry Myerson Aug 21 '11 at 13:03

It is a field, and you don't need to find an inverse for each element to prove that each element has an inverse. You can prove that if $\alpha$ is in the set then $\lbrace\,1,\alpha,\alpha^2,\alpha^3\,\rbrace$ is a linearly dependent set over the rationals, then deduce that $\alpha$ satisfies an equation of degree (at most) 3 over the rationals, then if $A\alpha^3+B\alpha^2+C\alpha+D=0$ you have $\alpha(A\alpha^2+B\alpha+C)=-D$, from which you can see an inverse to $\alpha$.

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Gerry, add a discussion about the nonvanishing of at least one of the coefficients (e.g., you don't see an inverse immediately if D = 0, but in that case A, B, or C is not 0 so...). –  KCd Aug 21 '11 at 1:16
    
@KCd, of course you're right, but I have to leave something for OP to do.... –  Gerry Myerson Aug 21 '11 at 1:18

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