Universal Property for Simple Field Extensions

Theorem
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 $\phi = \psi \big|_K$.

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:

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.

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 Evaluation Homomorphism 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 for 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.