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I'm making some exercises to prepare for my ring theory exam:

Is $\mathbb{Q}[α]=\{a+bα+cα^2 :a,b,c ∈ \mathbb{Q}\}$ with $α=\sqrt[3]{2}$ a field ?

If $(a+bα+cα^2)(a'+b'α+c'α^2)=1$, then (after quite some calculation and noticing that $α^3=2$ and $α^4=2α$): \begin{align*} aa'+2bc'+2cb'&=1 \\ ab'+ ba'+2cc'&=0 \\ ca'+bb'+ac' &= 0 \end{align*}

I'm not sure how to proceed, and if I'm heading in the right direction. Any help would be appreciated.

Something else I was thinking about, this ring I have seems to be isomorphic to:

$$\mathbb{Q}[X]/(X^3-2)$$

But this is not a maximal ideal, as it is contained in the ideal $(X^3,2)$. Would this be correct reasoning ?

Martin Sleziak
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Kasper
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    Related: http://math.stackexchange.com/q/294993/264 – Zev Chonoles Oct 20 '13 at 20:33
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    The ideal $(X^3,2)$ is the whole of the ring $\mathbb Q[X]$ (since it contains $2$, it contains $2\times \frac12 P(X)=P(X)$ for any polynomial $P$), so that does not stop $(X^3-2)$ being a maximal ideal. In fact, $(X^3-2)$ is a maximal ideal, since it is irreducible in $\mathbb Q[X]$. – John Gowers Oct 25 '13 at 10:33

3 Answers3

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One approach would be to consider $\mathbb{Q}[\alpha]$ as a $\mathbb{Q}$-vector space of dimension $3$.

Let $\beta = a + b\alpha + c\alpha^2$. Now consider the numbers $1$, $\beta$, $\beta^2$, and $\beta^3$. Because $\mathbb{Q}[\alpha]$ has dimension $3$, these are linearly dependent, so there are $u,v,w,x\in \mathbb{Q}$ with $u+v\beta + w\beta^2 + x\beta^3 =0$.

We can assume $u\neq 0$ (why?), so $\beta(v+w\beta+x\beta^2)=-u$, and we're almost done...

Andrew Dudzik
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  • This is quite clever, but I don't see how I could come up with something like this myself. – Kasper Oct 20 '13 at 21:38
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    @Kasper It's a standard trick in the subject—I remember that it was shocking to me when I first saw it. Basically, it's a form of pigeonhole argument, where you use finiteness properties of your module/field/whatever to generate relations out of thin air—a similar example would be the proof that a finite integral domain is a field, where you just try everything and prove that something has to work. – Andrew Dudzik Oct 20 '13 at 21:54
  • @Kasper But it's philosophically similar to anon's proof (if not identical) and also related to the general fact that algebraic numbers form a ring. Finally, it's somehow natural to think of multiplication by $\beta$ as a linear transformation of the vector space $\mathbb{Q}[\alpha]/\mathbb{Q}$, and what I am doing here is computing its characteristic polynomial in a somewhat sneaky way. In short: learn these things one at time, and eventually, when you know an area from a whole bunch of different perspectives, it'll be second-nature. – Andrew Dudzik Oct 20 '13 at 21:58
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There is a purely abstract way to do this in higher generality:

Let $K$ be a field and suppose $\alpha$ exists in an extension and is algebraic over $K$. Therefore we can say that $K[\alpha]\cong K[x]/f(x)$ where $f(x)$ is $\alpha$'s minimal polynomial over $K$. We want to show that any nonzero element $g(x)\in K[x]/f(x)$ is invertible, i.e. $\exists u,v$ s.t. $u(x)f(x)+v(x)g(x)=1$...

anon
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There is also a low-brow approach.

It will suffice that the three equations you wrote out always have a solution $a',b',c'$ for any $a,b,c$, except $0,0,0$.

This is a linear system of equations, with the corresponding matrix $$A = \begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{bmatrix}.$$ So, to check that there is a solution, you just need to check that $\det A \neq 0$. But $\det A = a^3 + 2b^3 + 4 c^3 - 6abc$, which is non-negative by the inequality between arithmetic and geometric means, and which would be $0$ only if $a^3 = 2b^3 = 4c^3$ - but this cannot happen for non-zero integers.

Jakub Konieczny
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