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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?

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    $\begingroup$ If $\alpha$ is algebraic, then $L[\alpha]$ isn't a polynomial ring. $\endgroup$ – Future Jan 29 '16 at 2:14
  • $\begingroup$ How are you certain of that? $\endgroup$ – user230283 Jan 29 '16 at 2:19
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    $\begingroup$ For me, it doesn't make sense to talk about "polynomial ring" when the indeterminate has an algebraic relation. What does $L[\alpha]$ even mean in this case? The natural definition would just coincide with $L(\alpha)$. Edit: Actually, is that what you want to show? That the field extension $L[x]/m_\alpha (x) = L(\alpha)$? $\endgroup$ – Future Jan 29 '16 at 2:22
  • $\begingroup$ Is there another way to do that? We have not seen this material in class so far. $\endgroup$ – user230283 Jan 29 '16 at 3:03
  • $\begingroup$ You mean the ring of polynomial expressions in $\alpha$. I don’t think it’s such a terrible solecism to refer to it the way you did, especially since your meaning was clear. $\endgroup$ – Lubin Jan 29 '16 at 5:01
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$\alpha \in K$ 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.

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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, $L[\alpha] = L(\alpha)$.

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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.

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