How to prove the diagram of ring homomorphisms conmutes?

Question

Let $D$ be an integral domain and $Q$ be its quotient field. Let $\varphi:D\to Q$ the inclusion (monomorphism) $\varphi(d)=[d,1]$. I need to prove that if $\mathbb{K}$ is any other field such that we have a monomorphism $\psi:D\to\mathbb{K}$, then exists an unique monophormism $\hat{\psi}:Q\to\mathbb{K}$ such that the diagram conmutes, i.e. $\psi=\hat{\psi}\circ\varphi$. In other words, any field $\mathbb{K}$ that contains $D$ must contain a "copy" of the quotient field $Q$ of $D$, in this meaning, $Q$ is the smallest field that contains $D$. Also, does this property characterize the quotient field $Q$ of $D$?

Answer 1

Suppose $\hat{\psi}$ exists such that $\hat{\psi}\circ\varphi=\psi$. Then, for every $a,b\in D$, with $b\ne0$, we must have $$ \hat{\psi}\left(\frac{a}{b}\right) =\hat{\psi}\left(\frac{a}{1}\right)\hat{\psi}\left(\frac{1}{b}\right) $$ Since $1/b=(b/1)^{-1}$, we can go on with the chain of equalities $$ =\hat{\psi}(\varphi(a))\bigl(\hat{\psi}(\varphi(b))\bigr)^{-1} =\psi(a)(\psi(b))^{-1} $$ Therefore $\hat{\psi}$ only depends on the action of $\psi$ on $D$, so we have proved uniqueness.

This also shows how $\hat{\psi}$ must be defined: $$ \hat{\psi}\left(\frac{a}{b}\right)=\psi(a)(\psi(b))^{-1} $$

You just have to prove this is well defined and a homomorphism.

First, it doesn't depend on the particular representation of $a/b$. Indeed, if $a/b=c/d$, then $ad=bc$ by definition, so $\psi(ad)=\psi(bc)$ and therefore $\psi(a)\psi(d)=\psi(b)\psi(c)$. Hence, as in $K$ nonzero elements have an inverse (here we're using that $\psi$ is injective) $$ \psi(a)(\psi(b))^{-1}=\psi(c)(\psi(d))^{-1} $$

Checking that $\hat{\psi}$ defined as above is a ring homomorphism should be easy enough.

I preferred the usual notation $a/b$ instead of $[a,b]$, but it's just a matter of substituting the symbols.

Answer 2

Uniqueness: We must have $\hat\psi([d,1])= \psi(d)$ and hence $$\tag1\hat\psi([x,y])=\hat\psi([x,1]/[y,1])=\hat\psi([x,1])/\hat\psi([y,1])=\psi(x)/\psi(y).$$ Existence: We need to show that $\hat\psi$ is well-defined by $(1)$. Indeed, if $[x,y]=[u,v]$ with $y,v\ne0$ then $xv-uy=0$, hence $\psi(x)\psi(v)=\psi(u)\psi(y)$ and (as $\psi$ is a monomorphism and so $\psi(y),\psi(v)$ are non-zero) $\psi(x)/\psi(y)=\psi(u)/\psi(v)$, as desired.

Morphism: $\hat\psi$ defined by $(1)$ is a compatible with multiplication: $$\hat\psi([x,y][u,v])=\hat\psi([xu,yv])=\frac{\psi(xu)}{\psi(yv)}=\frac{\psi(x)\psi(u)}{\psi(y)\psi(v)}=\frac{\psi(x)}{\psi(y)}\cdot\frac{\psi(u)}{\psi(v)}=\hat\psi([x,y])\hat\psi([u,v])$$and addition: $$ \begin{align}\hat\psi([x,y]+[u,v])&=\hat\psi([xv+uy,yv])\\&=\frac{\psi(xv+uy)}{\psi(yv)}\\&=\frac{\psi(x)\psi(v)+\psi(u)\psi(y)}{\psi(y)\psi(v)}\\&=\frac{\psi(x)}{\psi(y)}+\frac{\psi(u)}{\psi(v)}\\&=\hat\psi([x,y])+\hat\psi([u,v])\end{align}$$

Monomorphism: A ring homomoprhism with a field as domain is monomorph as soon as it is non-trivial. Here, $\hat\psi([1,1])=\psi(1)/\psi(1)=1\ne 0$.

---

As usual with universal properties, this characterizes $Q$ (together with $\phi$) up to unique isomorphism. If $Q'$ together with $\phi'\colon D\to Q'$ also has the described universal property, then this property grants us a unique morphism $\widehat{\phi'}\colon Q\to Q'$ with $\phi'=\widehat{\psi'}\circ \phi$ and a unique morphism $\hat{\psi}\colon Q'\to Q$ with $\phi=\hat\phi\circ\phi'$. The compositions $\widehat{\phi'}\circ \hat\phi\colon Q\to Q$ and $\hat\phi\circ \widehat{\phi'}\colon Q'\to Q'$ must coninced (again by uniqueness) with the respective identities, i.e., $\hat\psi$ is the claimed unique isomorphism.