If $\alpha$ is an algebraic element and $L$ a field, does the polynomial ring $L[\alpha]$ is also a field? If $\alpha$ is an algebraic element and $L \subset K$ are both field, does the polynomial ring $L[\alpha]$ is also a field?
I am trying to prove that the ring of fraction $L(\alpha)$ is equal to $L[\alpha]$. To do this, I use an exercise done in class : Let $K$ a field, $\alpha \in K$ and $L$ a subfield of $K$. Then $L(\alpha)$ is the smallest subfield of $K$ containing $L$ and $\alpha$.
Is anyone could help me?
 A: $\alpha \in K$ is algebraic over $L$ iff $L[\alpha]$ is finite-dimensional over $L$.
Take $\beta \in L[\alpha]$ and consider $\mu: x \mapsto \beta x$. Then $\mu$ is an $F$-linear transformation of $L[\alpha]$ which is injective if $\beta\ne0$ because $K$ is a field. Since $L[\alpha]$ is finite-dimensional over $L$, $\mu$ must be surjective. In particular, $1$ is in the image of $\mu$ and so $\beta$ is invertible. Thus, $L[\alpha]$ is a field.
A: I'm going to assume $\alpha \in K$ is algebraic over $L$ (otherwise we may have a transcendental extension, in which case $L[\alpha]$ is most definitely not the same thing as $L(\alpha)$).
Since $\alpha$ is algebraic over $L$, there is an irreducible polynomial $f = \sum_{i=0}^n a_i X^i$ in $L[X]$ with coefficients $a_0, \ldots, a_n$ in $L$ and minimal degree $n \in \mathbb{N}$. Additionally, $a_0 \neq 0$ (otherwise $f$ cannot be irreducible). We have $f(\alpha) = a_0 + \sum_{i=1}^n a_i \alpha^i = 0$, and therefore
$$a_0 = - \alpha \sum_{i=1}^n a_i \alpha^{i-1}.$$
Multiplying by $\frac{1}{a_0} \in L$ gives us that $- \frac{1}{a_0} \sum_{i=1}^n a_i \alpha^{i-1} = \alpha^{-1} \in L$. As a result, $L[\alpha] = L(\alpha)$.
A: There is a surjective ring homomorphism
$$
\text{ev}_\alpha \colon L[X] \to L[\alpha]
$$
that sends $p(X)$ to $p(\alpha)$.  Write $I$ for the kernel of this homomorphism - i.e., the set of all polynomials over $L$ that have a zero at $\alpha$.  By the First Isomorphism Theorem for Rings, we have an isomorphism
$$
L[X] / I \cong L[\alpha]
$$
(sending the coset $p(X) + I$ to $p(\alpha)$).
Since $k[\alpha]$ is certainly an integral domain (being a sub-ring of $K$), this means that $I$ is a prime ideal in $L[X]$.  But $L[X]$ is a principal ideal domain, and so every prime ideal in $L[X]$ is either the zero ideal or is maximal. However, since $\alpha$ is algebraic over $L$, there must be at least one non-zero $p$ in $I$.
Therefore, $I$ is a maximal ideal in $L[X]$ and so $L[\alpha]$ is a field.
