Does $K[\alpha]$ has an inverse of $\alpha$ if this element is algebraic?

Question

I was asking myself a question :

Let $K$ be a field and $K[\alpha]$ the smallest ring containing $K$ and $\alpha$. If $\alpha$ is algebraic, can we always find $\alpha^{-1}$ in $K[\alpha]$.

My thoughts were as following. I $p$ is the minimal polynomial of $\alpha$, than if we suppose moreover (do we have to???)

$gcd(x,p) = 1$

the gcd = 1 can be rewritten as :

$x f(x) + p(x) g(x) = 1$

$\alpha f(\alpha) + p(\alpha) g(\alpha) = 1$

so because $p(\alpha) = 0$

$f(\alpha) = \alpha^{-1}$

But can we not assume the gcd with $x$ is 1?

Answer 1

$K[\alpha]$ is a finite-dimensional vector space over $K$. The "multiply by $\alpha$" map is linear and injective. Therefore surjective.

For example, if $\alpha$ has minimal polynomial $$ 1+2\alpha-3\alpha^2+\alpha^3=0 $$ then divide by alpha $$ \frac{1}{\alpha}+2-3\alpha+\alpha^2=0 $$ and solve $$ \frac{1}{\alpha}=-2+3\alpha-\alpha^2 $$

Answer 2

Now let's see . . .

First of all, the method proposed by our OP roi_saumon appears to be correct in its essentials, though lacking in a proof that $\gcd(x, p(x)) = 1$; how may we show this is the case? Well, if

$d(x) \mid x, \; d(x) \mid p(x) \; \text{with} \; x, \; d(x), \; p(x) \in K[x], \tag 1$

there is some $q(x) \in K[x]$ with

$q(x)d(x) = x; \tag 2$

it then follows that

$\deg q(x) + \deg d(x) = \deg x = 1, \tag 3$

and thus either

$\deg q(x) = 0, \; \deg d(x) = 1 \; \text{or} \; \deg q(x) = 1, \; \deg d(x) = 0; \tag 4$

we may rule out the former case since, by (2), we then necessarily have

$d(x) = ax, \; q(x) = a^{-1}, \; 0 \ne a \in K; \tag 5$

but then $d(x) \mid p(x)$ implies the existence of

$t(x) = \displaystyle \sum_0^{\deg p - 1} t_i x^i \in K[x] \tag 6$

with

$p(x) = ax t(x) = \displaystyle \sum_0^{\deg p - 1} a t_i x^{i + 1}, \tag 7$

according to this equation,

$a \alpha t(\alpha) = p(\alpha) = 0, \tag 8$

whence if $\alpha \ne 0$ we have

$t(\alpha) = 0; \tag 9$

but

$\deg t(x) \le \deg p(x) - 1 < \deg p(x), \tag{10}$

and this contradicts the minimality of $p(x)$ with respect to $\alpha$; therefore, we may rule out the case $\deg q(x) = 0$, $\deg d(x) = 1$; on the other hand, with $\deg q(x) = 1$, $\deg d(x) = 0$ for every common divisor $d(x)$ of both $x$ and $p(x)$, it is certainly true that we may take

$\gcd(x, p(x)) = 1, \tag{11}$

and from this point our OP's approach to the problem is seen to be both rigorous and fruitful.

Another way to approach this problem, one which in fact develops a explicit formula for $\alpha^{-1}$, is based upon the observation that the minimal polynomial $p(x) \in K[x]$ of $\alpha$ must have a non-vanishing constant term. Otherwise, we may write

$p(x) = \displaystyle \sum_1^{\deg p} p_i x^i = x \sum_1^{\deg p} p_i x^{i - 1}, \tag{12}$

which implies

$\displaystyle \alpha \sum_1^{\deg p} p_i \alpha^{i - 1} = p(\alpha) = 0; \tag{13}$

under the assumption $\alpha \ne 0$, (13) additionally implies

$\displaystyle \sum_1^{\deg p} p_i\alpha^{i - 1} = 0, \tag{14}$

which asserts that $\alpha$ satisfies a polynomial of degree less that $\deg p(x)$, contradicting the minimality of $p(x)$; therefore

$p_0 \ne 0 \tag{15}$

and we have

$\displaystyle \sum_0^{\deg p} p_i \alpha^i = 0, \tag{16}$

which may be written

$\alpha \displaystyle \sum_1^{\deg p} p_i \alpha^{i - 1} = \sum_1^{\deg p} p_i \alpha^i = -p_0, \tag{17}$

or

$\alpha \displaystyle \sum_1^{\deg p} -\dfrac{p_i}{p_0} \alpha^{i - 1} = 1, \tag{18}$

which shows both that $\alpha^{-1}$ exists and is given by the formula

$\alpha^{-1} = -\displaystyle \sum_1^{\deg p} \dfrac{p_i}{p_0} \alpha^{i - 1} \in K[\alpha]. \tag{19}$