A homomorphism that extends to a place or an embedding

Question

I tried to learn commutative algebra and I found a problem to do the following:

Consider an integral domain $R$ and an algebraically closed field $M$. Let $\varphi_0$ be a homomorphism from $R$ to $M$, and let $F$ be a field that contains $R$. In this context, we can explore two possibilities for extending $\varphi_0$: either as an embedding $\varphi$ of $F$ into $M$, or as a place $\varphi$ of $F$ into $M\cup \{\infty\}$.

In this context, a place is defined as follows:

A place of a field $F$ is a map $\varphi$ into a set $M\cup \{\infty\}$, where $M$ is a field satisfying:

1. $\varphi(a+b)=\varphi(a)+\varphi(b)$
2. $\varphi(ab)=\varphi(a)\varphi(b)$
3. There are $a,b\in F$ such that $\varphi (a)=\infty$ and $\varphi(b)\ne 0,\infty$.

Also, the symbol $\infty$ has the following properties:

4. $x+\infty=\infty+x=\infty$ for all $x\in M$.
5. $x\cdot\infty=\infty\cdot x=\infty\cdot\infty=\infty$ for all $x\in M^\times$.
6. Neither $\infty+\infty$, nor $0\cdot\infty$ are defined.

Now, I try to figure out the following:

1. What is an example where $\varphi_0$ extends to an embedding?
2. How to prove the theorem that the extension leads always to an embedding or to a place?

I already found that if $\varphi(1)=1$ then $\varphi$ can be extended to a place of $F$. (S. Lang, Introduction to algebraic geometry (1973) Theorem 1, page 8) that was suggested by M. D. Fried and M. Jarden, Field Arithmetic (2008), third edition, Proposition 2.3.1, page 24.

Answer 1

Addressing the first question one can say the following (I assume that $\phi_0(1)=1$): The field $F$ contains the field $K$ of fractions of $R$. A necessary condition for $\phi_0$ to extend to an embedding of $F$ into $M$ is that $\phi_0$ be injective. In that case $\phi_0$ naturally extends to an embedding $\phi_1$ of $K$ into $M$. Now if the cardinality of a transcendence basis $T$ of the field extension $F/K$ is not bigger than the cardinality of a transcendence basis of the field extension $M/\phi(K)$ the homomorphism $\phi_1$ extends to a homomorphism $\phi_2$ of the rational function field $K(T)\subseteq F$ into $M$. Since the field extension $F/K(T)$ is algebraic and $M$ is algebraically closed the homomorphism $\phi_2$ extends to a homomorphism $\phi$ of $F$ into $M$.