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 $T = R \cup (S \setminus \operatorname{im}\phi)$.

Define $\theta : T \to S$ as follows:


 * If $x \in R$, then $\theta(x) = \phi(x)$
 * If $x \in (S \setminus \operatorname{im}\phi)$ then $\theta(x) = x$

We claim that $\theta$ is an isomorphism.

Injectivity: Let $\theta(x) = \theta(y)$.

Since $\theta(R) = \phi(R)$, we have $\theta(R) \cap ( S \setminus \operatorname{im}\phi) = \varnothing$.

Therefore either $x,y \in R$ or $x,y \in (S \setminus \operatorname{im}\phi)$.

If $x,y \in R$ then $\phi(x) = \phi(y)$, so $\theta$ is injective because $\phi$ is.

If $x,y \in (S \setminus \operatorname{im}\phi)$ then $\theta$ acts as the identity and $x = y$.

Surjectivity: Under $\theta$, $R$ surjects onto $\operatorname{im}\phi = \phi(R)$ by definition.

Moreover, $\theta(S \setminus \operatorname{im}\phi) = \operatorname{id}(S \setminus \operatorname{im}\phi) = (S \setminus \operatorname{im}\phi)$.

Therefore $\theta$ is surjective everywhere.

Now we endow $T$ with addition and multiplication: for $x,y \in T$, let


 * $x + y = \theta^{-1}(\theta(x) + \theta(y)),\quad xy = \theta^{-1}(\theta(x)\theta(y))$

so trivially we have


 * $\theta(x + y) = \theta(x) + \theta(y),\quad \theta(xy) = \theta(x)\theta(y)$

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.