How many monomorphisms are there $\mathbb{Q} \rightarrow \mathbb{C}$

Question

I was surprised as to how no sources online really took this particular monomorphism for an example while it seemed very common.

My thoughts are that it is only the identity map that exists, so $1$.

I am essentially mapping $p \in \mathbb{Q}$ to $a+bi \in \mathbb{C}$. But for it to be a monomorphism, I need $\phi(1)=1$(right?). Then I first have the map $\phi(p)=p+0i=p$

Now if I try to come up with a different monomorphism, namely a non-identity map, so if I have any $t \in \mathbb{Q}$, $t=\frac{p}{q}$ then $\phi'(t=\frac{p}{q})=\frac{\phi'(p)}{\phi'(q)}$. But even for this map, I must have the rule $\phi'(1)=1$. Then, $\phi'(p)=\phi'(1+...+1)=p\phi'(1)=p$ and in a similar manner we end up $\phi'$ mapping it to the same element in $\mathbb{C}$.

So $\phi'$ also becomes the identity element no matter what.

Am I right? If not, can someone tell me why and how to fix it please.

Answer 1

This should be observed from somewhat more general angle which is very important.

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First of all, $\mathbb Z$ is very special (unital) ring:

For any unital ring $R$ there is precisely one ring homomorphism (that preserves unity) from $\mathbb Z$ to $R$.

This is easy to prove: for any $f\colon\mathbb Z\to R$, we have $f(n) = nf(1) = n\cdot 1_R$.

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Secondly, $\mathbb Q$ is again a special field - it is field of fractions of the ring $\mathbb Z$. Now, fields of fractions are characterized by universal property:

For integral domain $R$, field $\operatorname{Frac}(R)$ is field of fractions of $R$ if there is ring inclusion $i\colon R\to \operatorname{Frac}(R)$ and if for any field $K$ and any ring monomorphism $f\colon R\to K$, there exists unique ring homomorphism $\bar f\colon \operatorname{Frac}(R)\to K$ such that $\bar f\circ i = f$.

That is, any inclusion of $R$ in some field factors through $\operatorname{Frac}(R)$, or, $\operatorname{Frac}(R)$ is the smallest field that contains $R$.

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How do we show this for $\mathbb Z$ and $\mathbb Q$?

Let $f\colon \mathbb Z\to K$ be monomorphism. Define $\bar f\colon \mathbb Q\to K$ as follows: $\bar f(\frac m n) = f(m)f(n)^{-1}$. It's easy to see that this is well defined if we note that $$\bar f\left(\frac{mk}{nk}\right) = f(mk)f(nk)^{-1} = f(m)f(k)f(k)^{-1}f(n)^{-1} = f(m)f(n)^{-1} = \bar f\left(\frac m n\right)$$ Also, let $g\colon \operatorname{Frac}(R)\to K$ such that $g\circ i = f$. Then, we have $$1_K = g(1) = g\left(n\cdot\frac 1 n\right) = g(n)g\left(\frac 1 n\right)\implies g\left(\frac 1 n\right) = g(n)^{-1}$$ $$g\left(\frac m n\right) = g\left(m\cdot\frac 1 n\right) = g(m)g\left(\frac 1 n\right) = g(m)g(n)^{-1} = f(m)f(n)^{-1} = \bar f\left(\frac m n\right)$$ This shows uniqueness of $\bar f$.

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So, after this long introduction, the answer is, as you say, only one ring homomorphism from $\mathbb Q$ to $\mathbb C$:

Let $f\colon\mathbb Q\to\mathbb C$. Then, $f(1) = 1$ and consequently $f(n) = n$, for all $n\in\mathbb Z$. Thus, $f$ extends (unique) inclusion of $\mathbb Z$ in $\mathbb C$, and thus is unique ring homomorphism from $\mathbb Q$ to $\mathbb C$, by $\mathbb Q$ being field of fractions of $\mathbb Z$.

Answer 2

If you're thinking about $\Bbb Q$ and $\Bbb C$ as rings, then yes, there is only one monomorphism.

$\phi(1)$ must be idempotent, and cannot be $0$ (because of the "mono-" in monomorphism, or alternatively, because some authors require ring homomorphisms to take $1$ to $1$), so it must be $1$. From there you can show that $\phi(p/q) = p/q$, by first showing that $\phi(1/n) = 1/n$ for any $n \in \Bbb Z$ via linearity of homomorphisms on $n/n = 1/n + \cdots + 1/n$, and then again use linearity on $p/q = 1/q + \cdots + 1/q$.