Theorem: Universal Property for Simple Field Extensions

Let $F / K$ be a field extension.
Let $\alpha \in F$ be an algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial over $\alpha$ over $K$.
Let $\psi : K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \restriction_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
Then $\psi\left({\alpha}\right)$ is a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi  \left({\mu_\alpha}\right)$ in $L$.
Then there exists a unique field homomorphism $\psi : K \left({\alpha}\right) \to L$ extending $\phi$ that sends $\alpha$ to $\beta$.

Proof:
Let $\psi: K \left({\alpha}\right) \to L$ be a homomorphism.
Let $\phi = \psi \big|_K$.
Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms.
For any $f = a_0 + \dotsb + a_n X^n \in K \left[{X}\right]$ we have:
{{begin-eqn}}
{{eqn | l = \psi \left({f \left({\alpha}\right)}\right)
      | r = \psi \left({a_0 + \cdots + a_n \alpha^n}\right)
      | c = 
}}
{{eqn | r = \phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) \psi \left({\alpha}\right)^n
      | c = by the homomorphism property, and the fact that $\phi = \psi \big\vert_K$
}}
{{eqn | r = \overline \phi \left({f}\right) (\psi\left({\alpha}\right))
      | c = 
}}
{{end-eqn}}
Since $\mu_\alpha \left({\alpha}\right) = 0$ and a Ring Homomorphism Preserves Zero, the above yields:
:$\overline \phi \left({f}\right) \left({\psi \left({\alpha}\right)}\right) = 0$
as required.
{{qed|lemma}}
Conversely let $L$ be a field.
Let $\phi: K \to L$ be a homomorphism.
Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
By Structure of Simple Algebraic Field Extension, there exists a unique isomorphism:
:$\Delta: K \left[{\alpha}\right] \to K \left[{X}\right] / \langle \mu_\alpha \rangle$
such that:
:$\Delta \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
and:
:$\Delta \big|_K$ is the identity mapping.
By Universal Property of Polynomial Ring there exists a unique homomorphism:
:$\chi: K \left[{X}\right] \to L$
such that:
:$\chi \left({X}\right) = \beta$
and:
:$\chi \big|_K = \phi$
By Universal Property of Quotient Ring there exists a unique homomorphism:
:$\psi: K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \to L$
such that:
:$\chi = \psi \circ \pi$
So we have the following diagram:
::$\begin{xy}\xymatrix@L+2mu@+1em {
 K \left[{X}\right] \ar[r]^*{\pi}
                    \ar@{-->}[rd]_*{\exists ! \chi}
&
 \dfrac {K \left[{X}\right]} {\left\langle{\mu_\alpha}\right\rangle} \ar@{-->}[d]^*{\exists ! \psi}
                                                                     \ar[r]^*{\Delta}
&
 K \left[{\alpha}\right]
\\ 
&
 L
&
}\end{xy}$
Now we have:
:$\chi \left({X}\right) = \beta$
and:
:$\pi \left({X}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
Therefore, because:
:$\chi = \psi \circ \pi$
we have:
:$\psi \left({X + \left\langle{\mu_\alpha}\right\rangle}\right) = \beta$
Also:
:$\Delta^{-1} \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
so by the above:
: $\psi \circ \Delta^{-1}\left({ \alpha }\right) = \beta$
: $\psi \circ \Delta^{-1} \big|_K =  \psi\big|_K = \phi$
: the choice of $\Delta, \psi$ is unique.
{{qed}}
Category:Field Extensions
Category:Universal Properties
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