Extension fields - definitions I'm having a little hard to get around the definition of an extension field. In Abstract Algebra and it's applications, the extension field $\mathbb{Q}(\sqrt{2})$ is defined to be $\mathbb{Q}(\sqrt{2}) = \{ a + b\sqrt{2} :a,b \in \mathbb{Q} \}$. But is this true in general? That is, can one write $\mathbb{Q}(\alpha) = \{ a + b\alpha :a,b \in \mathbb{Q} \}$, where $\alpha $ is algebraic over $ \mathbb{Q}$?
I suppose not, since for example $ \mathbb{Q}(\sqrt[3]{2})$ has a basis of $ \{1,\sqrt[3]{2},(\sqrt[3]{2})^2 \}.$ So is it correct to let
$$ \mathbb{Q}(\sqrt[3]{2}) = \{ a + b\sqrt[3]{2} + c (\sqrt[3]{2})^2 : a,b,c \in \mathbb{Q} \}$$
or in more general,
$$ \mathbb{Q}(\alpha) := \{ k_0 + k_1 \alpha + \dots k_{n-1} \alpha^{n-1} : k_i \in \mathbb{Q} \} $$
Best regards
Confused student
 A: The essential crux of the matter is the following . Let $\alpha$ be an algebraic element which is a root of some irreducible polynomial of degree $n$ .
We define a ring $\displaystyle F[\alpha]=\{\sum_{i=1}^{m}a_{i}\alpha^{i}, a_{i}\in F,m\in\Bbb{N}\}$. Now as $\alpha$ is a root of a polynomial of degree $n$ , we can express $\alpha^{n}$ as a linear combination of $1,\alpha,...,\alpha^{n-1}$. So $F[\alpha]=\{\sum_{i=1}^{n-1}a_{i}\alpha^{i}\,,a_{i}\in F\}$. Then we see that the fraction field of this ring is essentially equal to itself and hence it is a field and we say it to be $F(\alpha)$ . And this will have basis $\{1,\alpha,...,\alpha^{n-1}\}$ as a vector space over $F$.
Note :- There exists a unique monic polynomial of minimum degree such that $\alpha$ is a root and it is called the Minimal Polynomial of $\alpha$ .  It is an easy excercise(definiton chasing) to see that the minimal polynomial is irreducible and conversely a monic irreducible polynomial of which $\alpha$ is a root is automatically the Minimal Polynomial.
However if you want to know the formal construction then see below.
In general if $f(x)$ of degree $n$ is an irreducible polynomial in $F[x]$ and if $\alpha$ is some root of this polynomial then we have that $\frac{F[x]}{(f(x))}$ is a field as $F[x]$ is a PID and as $f(x)$ is irreducible we have that $(f(x))$ is a maximal ideal.
Now consider the projection map $\pi:F[x]\to \frac{F[x]}{(f(x))}$ .
Then for any polynomial $g(x)$ we can verify that $\pi(g(x))=g(\pi(x))$.
Hence $f(\pi(x))=\pi(f(x))=f(x)\mod f(x)= 0\in \frac{F[x]}{(f(x))}$.
Hence $\pi(x)$ is a root of this polynomial in the extension $\frac{F[x]}{(f(x))}$ of $F$ . Now it is easy to see that $\{1,\pi(x),...,\pi(x)^{n-1}\}$ is an $F-$basis for $\frac{F[x]}{(f(x))}$ as a vector space over $F$. This is due to Euclid's Division Algorithm as any polynomial $g(x)\mod f(x)=a_{0}+a_{1}x+...+a_{n-1}x^{n-1}\mod f(x)$ . Hence $\pi(g(x))=a_{0}+a_{1}\pi(x)+...+a_{n-1}(\pi(x))^{n-1}$.
Let $\alpha$ be a root of this polynomial. (Note that if degree of $f(x)\geq 2$ then $\alpha\notin F$).
Then we define $\displaystyle F[\alpha]=\{\sum_{i=1}^{m}a_{i}\alpha^{i}, a_{i}\in F,m\in\Bbb{N}\}$. Now as $\alpha$ is a root of a polynomial of degree $n$ , we can express $\alpha^{n}$ as a linear combination of $1,\alpha,...,\alpha^{n-1}$. So $F[\alpha]=\{\sum_{i=1}^{n-1}a_{i}\alpha^{i}\,,a_{i}\in F\}=F(\alpha)$
Then define the ring homomorphism known as the evaluation map $e_{\alpha}:F[x]\to F(\alpha)$ by $e_{\alpha}(g(x))=g(\alpha)$ . Then this has kernel $(f(x))$ and hence we get the isomorphism $\frac{F[x]}{(f(x))}\cong F(\alpha)$ and hence $\{1,e_{\alpha}(\pi(x)),...,(e_{\alpha}(\pi(x)))^{n-1}\}=\{1,\alpha,...,\alpha^{n-1}\}$ is an $F-$ basis for $F(\alpha)$
