Prove that, there exist uncountably many multiplicative maps from $\Bbb{Q}^\ast\to\Bbb{Z}$

elementary-set-theoryfunctionsintegers

First of all, I should define multiplicative map-
A map $f:\Bbb{Q}^\ast\to\Bbb{Z}$ is said to be multiplicative map if $f(ab)=f(a)f(b)\ \forall a,b\in \Bbb{Q}^\ast $, here $\Bbb{Q}^\ast=\Bbb{Q}\backslash\{0\}$.
Let us assume on contrary, it is countable, and let the countable collection of all multiplicative maps be written by the enumeration $f_1, f_2,\ldots , f_n,\ldots$
Now, I want to construct another multiplicative map $\psi:\Bbb{Q}^\ast\to\Bbb{Z}$ using the maps $f_1, f_2,\ldots , f_n,\ldots$ but $\psi\ne f_i\ \forall i\in\Bbb{N}$. But I can't construct such a map.
Is there any other method to solve this problem? Can anybody suggest me a proper way out?
Thanks for assistance in advance.

Best Answer

Note that $f(a)$ admits an inverse, or $f(a)=0$. So there are really only three options for $f(a)$, which are $-1,0,1$. Moreover, $f(1)\neq -1$.

Let $\Bbb P$ denote the set of primes $\{2,3,5,\dots\}$ and let $p_n$ denote the $n$th prime, starting with $p_0=2$.

For $A\subseteq\Bbb P$, define $f_A(1)=1, f_A(-1)=-1$ and $f_A(p)=1$ if and only if $p\notin A$, and $f_A(p)=-1$ otherwise. I claim that this is enough to determine $f_A$ entirely.

  • For every $k\in\Bbb N$, write $k$ into its decomposition into primes, then $f_A(k)$ is determined exactly by the values given.

  • For every $k\in\Bbb{Z\setminus N}$, define $f_A(k)=-f_A(-k)$. Since $f_A(-1)=-1$, this definition is still multiplicative.

  • Next for $k\in\Bbb Z\setminus\{0\}$ define $f_A(\frac 1k)=f_A(k)$.

  • Finally, we get $f_A(\frac nm)=f_A(n)\cdot f_A(\frac 1m)$.

Now simply prove that if $A\neq B$, then $f_A\neq f_B$. But that's easy. And observe that $\mathcal P(\Bbb P)$ is in fact uncountable.

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