4

I was wondering if you could find an abelian group and define two different actions of a field on it such that the resulting vector spaces are of different dimensions over the field.

I tried various extensions of $\mathbb Q$. Now, since the set $V$ is already an abelian group, the action of $\mathbb Z$, and therefore $\mathbb Q$, are already determined. So I tried $\mathbb Q(\alpha)$, but I got stuck here too: if $\alpha$ is algebraic, then its action is restricted to satisfy its polynomial, so the vector spaces would still be isomorphic. If $\alpha$ is transcendental, I tried having it act trivially, but that doesn't work since it's impossible to extend this to rational functions in $\alpha$, as is required.

Is what I'm trying to do even possible? If so, could I have some hints on an example?

Nishant
  • 9,155

3 Answers3

3

Let $F=k(T)$ and $E=k(T^2$), where $k$ is any field. Then $F$ is a one-dimensional vector space over itself and, being an algebraic field extension $F=E(T)\cong E[X]/(X^2-T^2)$, a twodimensional vector space over $E$. But $F\cong E$ via $T\mapsto T^2$, obviously, so we have endowed the abelian group $F$ with two distinct and nonisomorphic $F$-vector space structures.

Hagen von Eitzen
  • 374,180
3

Consider $\mathbb{R}$ and $\mathbb{R}^2$. As $\mathbb{Q}$-vector spaces they have the same dimension, so they are isomorphic as abelian groups a fortiori. Of course, they are not isomorphic as $\mathbb{R}$-vector spaces.

Zhen Lin
  • 90,111
2

Let $\sigma:F\to F$ be a field endomorphism such that $[F:\sigma(F)]=\infty$.

The "twisted polynomial ring" $F[x;\sigma]$ is the set of polynomials with coefficients written on the left of powers of $x$, and the coefficients are not assumed to commute with $x$. Instead, $xa:=\sigma(a)x$ for all $a\in F$.

Take a look at the ideal generated by $x$ in the ring $F[x;\sigma]/(x^2)$. For all intents and purposes, it is the set $Fx+(x^2)$.

On the left

the action of scalars looks like this: $(\alpha+ (x^2))\cdot(\gamma x+(x^2))=\alpha\gamma x+(x^2)$

Consequently, $Fx+(x^2)$ is $1$ dimensional as a left $F$ vector space.

On the right

the action of scalars looks like this: $(\gamma x+(x^2))\cdot(\alpha+ (x^2))=\gamma\sigma(\alpha) x+(x^2)$

This is very different from the left side. By hypothesis, a basis of $F$ over $\sigma(F)$ has infinitely many elements. Let's fix such a basis $\mathcal{B}=\{\gamma_i \mid i\in I\}$. Now consider the set of elements $\{\gamma_i x+(x^2)\mid i\in I\}$. A right $F$ linear combination equal to zero looks like this:

$0+(x^2)=\sum_{j=1}^n (\gamma_ix+(x^2))\cdot (\alpha_i+(x^2))=\sum_{j=1}^n \gamma_i\sigma(\alpha_i)x+(x^2)=(\sum_{j=1}^n \gamma_i\sigma(\alpha_i))x+(x^2)$

Comparing the coefficients of both sides, this says $\sum_{j=1}^n \gamma_i\sigma(\alpha_i)=0$, and by linear independence of the $\gamma_i$ over $\sigma(F)$, $\alpha_i=0$ for all $i\in I$. So what we have shown is that $\{\gamma_i x+(x^2)\mid i\in I\}$ form an $F$-linearly independent set when $F$ acts on the right. On the right, $Fx+(x^2)$ is an infinite dimensional $F$ vector space.

(Additional note: you could let $[F:\sigma(F)]=n<\infty$ also, but you wouldn't want $n=1$ of course.)

rschwieb
  • 153,510
  • This answer is too implicit for a casual reader (like me) to understand. I suppose the mentioned ideal comes equipped with two structures of $F$-vector space, probably one of which involves $\sigma$, but this could have been made explicit. Also I assume the dimension is different for the two cases, this also could have been made clear. Finally, both for the ideal generated by $x^2$, and then in the quotient the one generated by $x$, it would be nice to say which type of ideal (left, right, two-sided) is meant. – Marc van Leeuwen Mar 15 '17 at 09:04
  • @MarcvanLeeuwen Thanks for the feedback. It did leave too much to the reader, so I incorporated changes explaining. – rschwieb Mar 15 '17 at 13:59