Universal Property for Simple Field Extensions

Theorem
Let $F/K$ be a field extension and $\alpha \in F$ 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 $\phi = \psi \big|_K$ and $\overline{\phi}:K\left[{ X }\right] \to L\left[{ X }\right]$ the induced homomorphism of polynomial rings.

Then $\psi\left({ \alpha }\right)$ is a root of $\overline{\phi}(\mu_\alpha)$ in $L$.

Conversely let $L$ be a field, $\phi : K \to L$ a homomorphism, and $\beta$ a root of $\overline{\phi}(\mu_\alpha)$ 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.

For any $f = a_0 + \cdots + a_n X^n \in K\left[{ X }\right]$ 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, $\phi : K \to L$ a homomorphism, and $\beta$ a root of $\overline{\phi}\left({ \mu_\alpha }\right)$ in $L$.

By Structure of Algebraic Field Extension, there is a unique isomorphism $\Delta : K\left[{ \alpha }\right] \to K\left[{ X }\right] / \langle \mu_\alpha \rangle$ such that $\Delta\left({ \alpha }\right) = X + \langle \mu_\alpha \rangle$ and $\Delta \big|_K$ is the identity.

By Evaluation Homomorphism there is 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 is a unique homomorphsim $\psi : K\left[{ X }\right] / \langle \mu_\alpha \rangle \to L$ 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 $\pi\left({ X }\right) = X + \langle \mu_\alpha \rangle$.

Therefore because $\chi = \psi \circ \pi$ we have $\psi\left({ X + \langle \mu_\alpha \rangle }\right) = \beta$.

Also $\Delta^{-1}\left({ \alpha }\right) = X + \langle \mu_\alpha \rangle$, 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.