**Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.8**

Solution: Call the additive and multiplicative identities of $\mb F$ $0_{\mb F}$ and $1_{\mb F}$ respectively. Define $n_{\mb F}$ to be the sum of $n$ $1_{\mb F}$’s. So $$n_{\mb F}=1_{\mb F}+1_{\mb F}+\cdots+1_{\mb F}$$ ($n$ copies of $1_{\mb F}$).

Define $-n_{\mb F}$ to be the additive inverse of $n_{\mb F}$. Since ${\mb F}$ has characteristic zero, if $n\not=m$ then $n_{\mb F}\not=m_{\mb F}$.

For $m,n\in\Z$, $n\not=0$, let $\left(\dfrac{m}{n}\right)_{\mb F}=m_{\mb F}\cdot n_{\mb F}^{-1}$. Since ${\mb F}$ has characteristic zero, if $\dfrac{m}{n}\not=\dfrac{m’}{n’}$ then $$\left(\dfrac{m}{n}\right)_{\mb F}\not=\left(\dfrac{m’}{n’}\right)_{\mb F}.$$ Therefore the map $\dfrac mn\mapsto\left(\dfrac{m}{n}\right)_{\mb F}$ gives a one-to-one map from $\mb Q$ to $\mb F$.

Call this map $h$. Then $h(0)=0_{\mb F}$, $h(1)=1_{\mb F}$ and in general $h(x+y)=h(x)+h(y)$ and $h(xy)=h(x)h(y)$. Thus we have found a subset of $\mb F$ that is in one-to-one correspondence to $\mb Q$ and which has the same field structure as $\mb Q$.