Definition 3

Let $K$ and $L$ be two fields. Every morphism $f$ of $K$ to $\tilde{L}$ (for addition and multiplication) such that $f(1) = 1$ is called a place of $K$ with values in $L$.

In other words, if $x$ and $y$ are elements of $K$ and $f(x) + f(y)$ (resp. $f(x)f(y)$) is defined, then $x + y$ (resp. $xy$) is defined and

As $\infty + \infty$ is not defined, neither is $f(\infty) + f(\infty)$, which shows that

Similarly, since $0 \cdot \infty$ is not defined, neither is $f(0)f(\infty)$, which, by virtue of (6), implies

On the other hand

If $f(a)f(a^{-1})$ is defined, $aa^{-1}$ is defined and hence is equal to 1; then $f(a)f(a^{-1}) = f(1) = 1$, which proves (8) in this case. If $f(a)f(a)^{-1}$ is not defined, then, either $f(a) = 0$ and $f(a^{-1}) = \infty$ or $f(a) = \infty$ and $f(a^{-1}) = 0$ and (8) still holds.

Similarly it can be shown that

From formulae (8) and (9) it follows that $f$ is also a morphism for the laws of composition $(x, y) \mapsto x - y$ and $(x, y) \mapsto xy^{-1}$.

For $x \in \tilde{K}$, $f$ is called finite at $x$ if $f(x) \neq \infty$; this implies $x \in K$ by (6).

If $f : K \to \tilde{L}$ is a place, $f(K)$ is a subset of $\tilde{L}$ which is stable for the laws of composition $(x, y) \mapsto x + y, (x, y) \mapsto x - y, (x, y) \mapsto xy$ and $(x, y) \mapsto xy^{-1}$ and which contains 1. If E is the set of finite elements of $f(\tilde{K})$, E is a subfield of L and $f(\tilde{K}) = \tilde{E}$. By an abuse of language E is called the valuefield off.

The composite mapping of two places is a place.

Let F be an isomorphism of a field K onto a subfield of a field L; let us extend $f$ to $K$ by setting $f(\infty) = \infty$. Thus we obtain a place of K with values in L which is called trivial and which is often identified with the isomorphism $f$.