4

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

Oskar
  • 812
  • 3
    It depends on $\alpha$. More precisely, it depends on the degree of its minimal polynomial $m(x)\in\Bbb{Q}[x]$. The latter recipe is correct, if $\deg m(x)=n.$ – Jyrki Lahtonen May 30 '22 at 10:43
  • 4
    It is a common misunderstanding to think that $\Bbb{Q}(\alpha)={a+b\alpha\mid a,b\in\Bbb{Q}}$. It originates from the fact that the first examples of extension fields are often quadratic. Should $\alpha$ be transcendental, then the meaning of $\Bbb{Q}(\alpha)$ is even more complicated. The definition of $\Bbb{Q}(\alpha)$ is the smallest extension field of $\Bbb{Q}$ containing $\alpha$. – Jyrki Lahtonen May 30 '22 at 10:46
  • 2
    @JyrkiLahtonen: As you know well, to be precise one even has to specify in what ("super-")extension we take such smallest extension field. In the beginner's examples, it's tacitly assumed everything happens inside $\mathbb R$ and $\mathbb C$, and then later people get confused when they're told an expression like $\mathbb Q_p(\sqrt[3]{5})$ is not well-defined, cf. https://math.stackexchange.com/q/4231699/96384 and the shenanigans in https://math.stackexchange.com/q/3258571/96384 and https://math.stackexchange.com/a/4007515/96384. MAybe OP need not care about these subtleties (yet). – Torsten Schoeneberg May 30 '22 at 15:52

1 Answers1

3

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 $F(\alpha)$

Mr.Gandalf Sauron
  • 13,906