Universal Property for Simple Field Extensions

Definition

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 $\psi : K(\alpha) \to L$ be a homomorphism, let $\phi = \psi \big|_K$ and $\overline{\phi}:K[X] \to L[X]$ the induced homomorphism.

Then $\psi(\alpha)$ 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 $\psi : K(\alpha) \to L$ extending $\phi$ that sends $\alpha$ to $\beta$.

Proof

First let $\psi : K(\alpha) \to L$ be a homomorphism, $\phi = \psi \big|_K$ and $\overline{\phi}:K[X] \to L[X]$ the induced homomorphism.

For any $f = a_0 + \cdots + a_n X^n \in K[X]$ we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \psi(f(\alpha))\)	\(=\)	\(\displaystyle \psi(a_0 + \cdots + a_n \alpha^n)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi(a_0) + \cdots + \phi(a_n) \psi(\alpha)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the homomorphism property, and the fact that $\phi = \psi \big\vert_K$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \overline{\phi}(f)(\psi(\alpha))\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Since $\mu_\alpha (\alpha) = 0$ and a ring homomorphism preserves zero, the above yields $\overline{\phi}(f)(\psi(\alpha)) = 0$ as required.

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

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

By Evaluation Homomorphism there is a unique homomorphism $\chi : K[X] \to L$ such that $\chi(X) = \beta$ and $\chi \big|_K = \phi$.

By Universal Property for Quotient Ring there is a unique homomorphsim $\psi : K[X] / \langle \mu_\alpha \rangle \to L$ such that $\chi = \psi \circ \pi$.

So we have the following diagram:

Now we have $\chi(X) = \beta$ and $\pi(X) = X + \langle \mu_\alpha \rangle$.

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

Also $\Delta^{-1}(\alpha) = X + \langle \mu_\alpha \rangle$, so by the above:

• $\psi \circ \Delta^{-1}(\alpha) = \beta$
• $\psi \circ \Delta^{-1} \big|_K = \psi\big|_K = \phi$
• The choice of $\Delta,\psi$ is unique.

$\blacksquare$