Surgery for Rings
Theorem
Let $R$ and $S$ be commutative rings with unity, and $\phi: R \to S$ a ring monomorphism.
Then there is a ring $T$ isomorphic to $S$ that contains $R$ as a subring.
Proof
Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
- If $x \in R$, then $\map \theta x = \map \phi x$
- If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
Injectivity: Let $\map \theta x = \map \theta y$.
Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$.
Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$.
If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is.
If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$.
Surjectivity: Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition.
Moreover:
- $\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$
where $I$ denotes the identity mapping.
Therefore $\theta$ is surjective everywhere.
Now we endow $T$ with addition and multiplication: for $x, y \in T$, let
- $x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$
- $x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$
so trivially we have
- $\map \theta {x + y} = \map \theta x + \map \theta y$
- $\map \theta {x y} = \map \theta x \, \map \theta y$
Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms.
Thus $T$ is a ring isomorphic to $S$ containing $R$ as a subring.
$\blacksquare$