For $F[\alpha]$ as a finite field extension of $F$, is $1/\alpha \in F[\alpha]$?

Question

Is the multiplicative inverse of $\alpha$, that is, (1/$\alpha$) contained within $F[\alpha]$? Thanks in advance.

Answer 1

If when you wrote "$F[\alpha]$ as a finite field extension of $F$" you meant that $F[\alpha]$ is a field containing $F$, then surely it has the inverse of $\alpha$: it is a field.

But perhaps you did not realize that "field extension" meant this. That's OK: we can deal with this.

In general, when you have two fields $K\supseteq F$, and $\alpha\in K\setminus{\{0\}}$, $F[\alpha]$ does not need to be a field... but at least it is a domain.

If you then add that $F[\alpha]$ is a finite dimensional vector space over $F$, then yes, $F[\alpha]$ is a field. So I guess this was the intention of the question: if $F[\alpha]$ is finite dimensional over $F$, then $F[\alpha]$ is a field.

Answer 2

If $\alpha\neq 0$, then the map $x \mapsto \alpha x$ is an injective linear transformation and so is surjective. This means that $1$ is in the image, which gives you an inverse for $\alpha$.

Answer 3

Well, if you suppose $F[\alpha]$ is a field (contained in some big extension of $F$), this means that $\alpha$ must be algebraic over $F$, i.e. there exists $P\in F[X]$ such that $P(\alpha)=0$ and $P\neq 0$ (for instance $\mathbb{Q}[\pi]$ is not a field). Writing the polynom $P$ :

$$P=a_nX^n+...+a_1X+a_0 $$

one sees that $P$ can be chosen such that $a_0\neq 0$ ($P$ can be chosen irreducible), because $P(\alpha)=0$ then :

$$0=a_n\alpha^n+...+a_1\alpha+a_0 $$

Finaly :

$$\alpha(a_n\alpha^{n-1}+...+a_1)\times(-a_0)^{-1}=1 $$

That's how you justify (given a big extension $K$ containing $\alpha$ and $F$) :

$$F[\alpha]\text{ is a field}\Leftrightarrow dim_F(F[\alpha])<\infty\Leftrightarrow \alpha\text{ is algebraic}$$

Answer 4

Let's assume that $\alpha \not\in F$ (otherwise $F[\alpha] = F$, which is a finite extension, but not very interesting). In particular, this forces $\alpha \neq 0$.

This means $F[\alpha]$ is a finite-dimensional vector space over $F$. In particular, the set:

$\{\alpha^k: k \in \Bbb N\}$ is linearly dependent, since it is infinite.

Choose $n$ to be the smallest positive integer such that $\{1,\alpha,\dots,\alpha^n\}$ is linearly dependent. This means there exists $c_0,\dots,c_{n-1} \in F$ such that:

$c_0 + c_1\alpha + \cdots +c_{n-1}\alpha^{n-1} + \alpha^n = 0$

I claim $c_0 \neq 0$, for if so, we have:

$\alpha(c_1 + c_2\alpha +\cdots + c_{n-1}\alpha^{n-2} + \alpha^{n-1}) = 0$.

Since $\alpha \neq 0$, we have $c_1 + c_2\alpha +\cdots + c_{n-1}\alpha^{n-2} + \alpha^{n-1} = 0$, contradicting our choice of $n$.

Since $c_0 \neq 0$, we have:

$1 = -\dfrac{1}{c_0}(c_1 + c_2\alpha +\cdots + c_{n-1}\alpha^{n-2} + \alpha^{n-1})\alpha$

that is: $\dfrac{1}{\alpha} = -\dfrac{1}{c_0}(c_1 + c_2\alpha +\cdots + c_{n-1}\alpha^{n-2} + \alpha^{n-1}) \in F[\alpha]$