if $a$ is algebraic, then $k[a] = k(a)$

Question

let $K/k$ be a field extension and $a\in K$ algebraic over k. Prove that the subalgebra $k[a]$ is a field and that $k[a] = k(a)$ where $k(a)$ is the monogenic extension, i.e. the smallest subfield of $K$ containing $k$ and $a$.

I know that $a$ is algebraic on $k[a]$ since $k\subset k[a]$. I don't see how to go from there. Thank you.

Answer 1

Since $x$ is algebraic, it satisfies an irreducible polynomial $p\in k[X]$, and

$$k[x]\cong k[X]/\langle p \rangle,$$

which is a field since the ideal $\langle p \rangle$ is maximal. Now clearly $k[x]\subseteq k(x)$, but also $k(x)\subseteq k[x]$ since $x^{-1}\in k[x]$.

EDITED:

The $k$-isomorphism between the $k$-algebras above is seen as follows:

Let $d$ be the degree of $p$. Then $B:=\{1,x,\dots,x^{d-1}\}$ is a $k$-linearly independent set inside $k[x]$ (otherwise the degree of $p$ would be less than $d$, since it is irreducible), and since $x^d$ is a linear combination of the vectors in $B$, we get that $B$ also spans $k[x]$, i.e., $B$ is a basis of the $k$-vector space $k[x]$. On the other hand, $K[X]/\langle p \rangle$ is clearly spanned by $B':=\{1,\overline{X},\ldots,\overline{X^{d-1}}\}$, where the overline indicates the canonical projection, and which is a basis by the same reasoning as before.

Let $\varphi:k[x]\rightarrow K[X]/\langle p \rangle$ be the linear map defined on $B$ by $\varphi(x^i):=\overline{X^i}$ (and extended by linearity). By construction, we see that $\varphi$ is in addition a homomorphism, since $x^k$ will go to $\overline{X^k}$ for every $k\in\mathbb{N}$ precisely because $p$ is the minimal polynomial of $x$, so that $\overline{X^k}$ satisfies the same linear combination on $B'$ as $x^k$ does on $B$.

Now $\varphi$ is bijective, since it is a linear map between bases of the same finite cardinal.

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For your future studies, this happens to be a particular case of the more general instance of the existence of free algebras: every $k$-algebra $A$ of "kind" $T$ (satisfying the identities in $T$) is isomorphic to the quotient of the free $k$-algebra of kind $T$ with the same number of generators as $A$ by the ideal $I(A)$ of identities satisfied by all the elements of $A$, where the free algebra is built from the most general "polynomials" which satisfy the identities $T$. So, for example, every associative and commutative algebra generated by one element is isomorphic to $k[X]/I$, where $I$ is the ideal of identities satisfied by that element, in this case, its minimal polynomial; also, every associative algebra generated by two elements such that $x^2=y-1$ (and not satisfying further relations) is isomorphic to $k\langle X,Y\rangle/I$, where $XY\neq YX$ and $I:=\langle X^2-Y+1\rangle$.

Answer 2

This is only a variant of Jose Brox' answer.

$k[a] \subset k(a)$ is obvious because the $k$-vector space $k[a]$ is generated by the powers $a^i$ which belong to $k(a)$.

It is well-known that $k(a)$ is a $k$-vector space with basis $1,a,..,a^{n-1}$, where $n$ is the degree of an irreducible polynomial with root $a$. This implies $k(a) \subset k[a]$.

Answer 3

To show that $k[a] \subset K$ is a field, we must needs establish that any $0 \ne y \in k[a]$ is possessed of a multiplicative inverse $y^{-1} \in k[a]$. Now since $a \in K$ is algebraic over $k$,

$n = [k[a]:k] < \infty, \tag 1$

from which it follows that a linear dependence exists between the first $n$ powers of $y$, that is, between

$1, \; y, \; y^2, \; \ldots, y^{n - 1}, y^n; \tag 2$

that is, there are $\alpha_i \in k$, $0 \le i \le n$, such that $y$ satisfies the polynomial

$\Upsilon(t) = \displaystyle \sum_0^n \alpha_i t^i \in k[t], \tag 3$

viz.,

$\Upsilon(y) = \displaystyle \sum_0^n \alpha_i y^i = 0; \tag 4$

this must be so since the subspace spanned by he $y^i$, $0 \le i \le n$, is of dimension at most $n = [k[a]:k]$. Now since $y$ satisfies a polynomial $\Upsilon(t) \in k[t]$ of degree at most $n$, we may affirm the existence of a polynomial

$\mu(t) \in k[t] \tag 5$

of minimal degree amongst all polyhnomials in $k[t]$ satisfied by $y$:

$\mu(y) = 0, \tag 6$

and

$\forall \Theta(t) \in k[t], \; \Theta(y) = 0 \Longrightarrow \deg \mu(t) \le \deg \Theta(t). \tag 7$

Now if $\mu(t)$ is such a polynomial,

$\mu(t) = \displaystyle \sum_0^{\deg \mu} \mu_i t^i, \; \mu_i \in k, \; 0 \le i \le \deg \mu(t), \tag 8$

I claim that

$\mu_0 \ne 0; \tag 9$

for if not, then

$\mu(t) = t \displaystyle \sum_1^{\deg \mu} \mu_i t^{i - 1}, \tag{10}$

and now (6) yields

$y \displaystyle \sum_1^{\deg \mu} \mu_i y^{i - 1} = \mu(y) = 0; \tag{11}$

since $y \ne 0$,

$\displaystyle \sum_1^{\deg \mu} \mu_i y^{i - 1} = 0, \tag{12}$

which shows $y$ satisfies a polynomial of degree less than that of $\mu(t)$; since by hypothesis this is impossible we conclude that $\mu_0 \ne 0$ and thus

$y \displaystyle \sum_1^{\deg \mu} \mu_i y^{i - 1} = \sum_1^{\deg \mu} \mu_i y^i = -\mu_0 \ne 0, \tag{13}$

or

$y \displaystyle \sum_1^{\deg \mu} -\dfrac{\mu_i}{\mu_0} y^{i - 1} = 1, \tag{14}$

which shows that

$y^{-1} = \displaystyle \sum_1^{\deg \mu} -\dfrac{\mu_i}{\mu_0} y^{i - 1} \in k[a]; \tag{15}$

we thus see that every $y \in k[a]$ is invertible and therefore that $k[a] \subset K$ is a field.

We need to show that $k(a) = k[a]$. Since $k \subset k(a)$ and $a \in k(a)$, we see that any $p(a) \in k[a]$, where $p(t) \in k[t]$, satisfies $p(a) \in k(a)$ since $k(a)$ is closed under the ring axioms; thus $k[a] \subset k(a)$; but we have just seen that $k[a]$ is itself a field, so since $k(a)$ is the smallest field (with respect to set inclusion "$\subset$") containing $k$ and $a$, we must have $k(a) = k[a]$; and we are done.