Is there a countable field such that the multiplicative group of that field is isomorphic to $(\mathbb{Z},+)$?
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asked Oct 5, 2017 at 8:37
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1$\begingroup$ No. If the field has characteristic 0, then it contains a copy of $\mathbb{Q}$, therefore its multiplicative group contains $\mathbb{Q}^*$. If the field has positive characteristic $p$, then it contains a copy of $\mathbb{F}_p$, therefore it has an invertible element $x$ such that $x^p = 1$. $\endgroup$– M.G.Commented Oct 5, 2017 at 8:50
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1$\begingroup$ Google says no: math.stackexchange.com/questions/753437/… $\endgroup$– DirkCommented Oct 5, 2017 at 8:50
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2$\begingroup$ No. In characteristic $\neq 2$, $-1$ has order 2. In characteristic 2, either $K$ is algebraic over $F_2$ and hence $K^*$ is torsion, or it has a transcendental element $x$. Since $x$ and $x+1$ freely generate a copy of $\mathbf{Z}^2$ in $K^*$, we conclude. $\endgroup$– YCorCommented Oct 5, 2017 at 8:52
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2$\begingroup$ @July the only $p$-th root of 1 in char $p$ is 1, so your argument does not work. Using the copy of $F_p$ still works, but only for $p\neq 2$. $\endgroup$– YCorCommented Oct 5, 2017 at 8:53
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$\begingroup$ @Ycor: oops, you are right of course about $p\neq 2$. Also, I meant $x^{p-1}=1$. $\endgroup$– M.G.Commented Oct 5, 2017 at 9:01
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1 Answer
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I’m going to say no. You want the non-zero elements to be $\cdots,x^{-1},1,x,x^2,\cdots$ where $x^k \neq 1$ for $k \neq 0.$
If the characteristic is a finite number $p$ then $1+x=x^j$ for some $j.$ If $j \gt 1$ this only allows $p^j$ elements. In case $j=-t \lt 0$ we have $x^t+x^{t+1}=1$ and $p^{t+1}$ elements.
In case of infinite characteristic every rational number is $x^j$ for a unique $j.$ In case that is not convincing enough, $2=x^j$ makes the field $\mathbb{Z}[\sqrt[j]{2}]$ which does not have the desired property, even for $j=\pm1.$
answered Oct 5, 2017 at 9:30