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Let $K$ be a field extension of the field $F$. Then an element $\alpha$ of $K$ is said to be

$(i)$ a root of a polynomial $f(x) = a_0 + a_1x +···+ a_nx^n$ in $F[x]$ iff $f (\alpha) = a_0 +a_1\alpha +···+ a_n\alpha_n = 0$;

$(ii)$ algebraic over $F$ iff $\alpha$ is a root of some non-null polynomial $f(x)$ in $F[x]$

Definition 1: Let $K$ be a field extension of the field $F$. Then $K$ is said to be an algebraic extension over $F$ iff every element of $K$ is algebraic over $F$.

Definition 2: A simple extension $F(\alpha)$ is said to be an algebraic over $F$ according to whether the element $\alpha$ is algebraic over $F$.

Here comes my question. To check $F(\alpha)$ over $F$ is algebraic or not, from definition 1 we have to check every element which is of the form $\frac{f(\alpha)}{g(\alpha)}$ where $f(x),g(x) \in F[x]$ and $g(\alpha) \neq 0$, is root of a some non null polynomial in $F[x]$. But from definition 2, to check algebraic or not we just have to check $\alpha$ is algebraic or not. What is the guarantee that if $\alpha$ is algebraic then other ratios of polynomial s evaluated at $\alpha$ will also be root of some other non null polynomial in $F[x]$?

Why is that? How definition 2 implies definition 1? Any help is appreciated thanks.

Alexander
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    I think this is fairly nontrivial. The easiest proof is probably checking that $\alpha\in F$ is algebraic iff $F[\alpha]$ is a finite-dimensional $F$-vector space, and observe that for any $x\in F[\alpha]$, we have $F[x]\subset F[\alpha]$ so $x$ is algebraic over $F$ as well. – Kenta S Jun 16 '22 at 07:04

2 Answers2

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If $F$ is a field and $K$ is a field extension of $F$, then the sum, difference, product, and quotient of two elements of $K$ which are algebraic over $F$ is still algebraic over $F$. So, if $\alpha$ is algebraic, then, for any rational fraction $\frac{f(x)}{g(x)}$ with coefficients in $F$, $\frac{f(\alpha)}{g(\alpha)}$ is algebraic too.

José Carlos Santos
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One possible approach is to observe that $F(\alpha)$ is a finite dimentional vector space over $F$. Note that you may start with proving that $F(\alpha)$ is actually same with $F[\alpha]$. This is an important observation since $F(\alpha)$ is by definition consists of a rational expression of elements of $F[\alpha]$ and the difficulties you have accountered vanishes.

To see $F[\alpha] = F(\alpha)$ is not that hard. How about to start with showing that $F[\alpha]$ is a field? To see this pick an element $\beta$ in there and consider 'vertors' $1, \beta, \beta^2, \ldots $ and make use of finiteness of the dimension of the vector space. This also provides that being algebraic of $F[\alpha]$.

seoneo
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