Show that the homomorphism from $K[X]$ to $L'$ which takes $X$ to $y$ induces an isomorphism of $L$ with $K(y)$.

Question

Let $K$ be a field, $F \in K[X]$ an irreducible monic polynomial of degree $n > 0$.

Suppose $L'$ is a field extension of $K, \; y \in L'$ such that $F(y) = 0$. Show that the homomorphism from $K[X]$ to $L'$ which takes $X$ to $y$ induces an isomorphism of $L$ with $K(y)$ where $L = K[X]/(F)$.

My solution: Let $\phi : K[X] \rightarrow L'$ s.t. $X \mapsto y$. We can see that $\phi (K[X]) = K[y] \subset K(y)$. So, we extend this map to $\phi ' : K[X] \rightarrow K(y) $ where it is easy to see that $(F) \subset ker(\phi')$. I don't know how to proceed.

Answer 1

Since $y$ is algebraic over $K$, we have that $K[y]=K(y)$. So your extension map is the identity map.

Since $ker(\phi)$ is an ideal in $K[X]$, which is a PID, because $K$ is a field, we have that $ker(\phi)=(g)$ for some $g \in K[X]$. We get from $(F) \subset (g)$, that $g$ divides $F$ in $K[X]$.

Notice that we have $g(y)=\phi(g)=0$, since $g \in ker(\phi)$. Since $F$ is irreducible, it follows that $F=g$. So we get the result by applying the isomorphism theorem.

Answer 2

So we're given $K\subseteq L'$ a field extension and irreducible non-constant $F\in K[x]$ with root $y\in L'$. Map $\varphi : K[x] \rightarrow L'$ by $\varphi (a) = a$ for $a\in K$ and $\varphi(x) = y$. Since $K$ is a field, $K[x]$ is a PID, so $F$ is prime, ('irreducible = prime' in UFDs, and PID $\implies$ UFD) and $(F)$ is maximal.

Then as $\varphi (F) = 0$, $(F)\subset ker \varphi$, and because it is maximal, $(F) = ker \varphi$. Then $K[x]/(ker\varphi) = K[x]/(F) \cong Im\varphi = H$, a subfield of $L'$. Clearly $K\subset H$ since $K = \varphi (K) \subset H$, and also $y = \varphi (x) \in H$.

By definition of $K(y)\subset L'$, that is, the smallest subfield of $L'$ containing $K$ and $y$, $K(y) \subseteq H$. But also, by definition of our map, all elements of $H$ look like $a_0 + a_1 y+ \cdots + a_n y^n$, with $a_k \in K$, which are also in $K(y)$, so $H\subseteq K(y)$.