$L \cong K(\sqrt{x}) \cong K[X]/(X^2-x)$ for $x \in K$

Question

I am trying to prove that if $L$ is a quadratic extension of a field $K$ with characteristic differing from 2, then $L \cong K(\sqrt{x}) \cong K[X]/(X^2-x)$ for $x \in K$.

I tried making a isomorphism from $K(\sqrt{x})$ to $K[X]/(X^2-x)$ given that $X^2-x=0$, so $f(X)^2-f(x)=0$, by using the abc-formula I got 1 and 0. These both go to 0 in $f(X)^2-f(x)=0$, but then the function would not be bijective.

Could anyone help me with a method that should work in order to prove these isomorphisms?

Answer 1

Let $L$ be a quadratic extension of $K$. Then $K(\alpha)=L$ for all $\alpha\in L\ \backslash \ K$ (because $\alpha$ is an algebraic element with degree $2$). We have to prove that exists $x\in K$ such that $L = K(\alpha) = K(\sqrt{x})$.

Take the minimum polinomial $\mu_\alpha(X) = X^2 + bX + c$ of $\alpha$. Since $char K\neq 2$ we have: $$ \alpha = \dfrac{-b+\sqrt{b^2-4c}}{2} $$ Then $K(\alpha)=K(\dfrac{-b+\sqrt{b^2-4c}}{2}) = K(\sqrt{\Delta})$. Take $\Delta=x$ and you are done.

The last isomorphism is given by the evaluation map: \begin{gather} K[X]\longrightarrow K[\sqrt{x}]\\ X \longmapsto \sqrt{x} \end{gather} That is surjective and the kernel is $(X^2 - x)$ that is maximal because $x$ is not a square in $K$.

Answer 2

You map $\sqrt{x}\longmapsto X$, and there is no other element of $K(\sqrt{x})$ which maps to $X$, so your mapping is bijective.