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.

$\blacksquare$