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.


marked as duplicate by Jyrki Lahtonen abstract-algebra Sep 21 '18 at 3:11

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    $\begingroup$ Isn't any $x \in k$ algebraic over $k$? Or do you mean something else? $\endgroup$ – Paul Frost Sep 20 '18 at 15:15
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    $\begingroup$ yes x is supposed to be algebraic over k $\endgroup$ – PerelMan Sep 20 '18 at 15:17
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    $\begingroup$ You should clarify what you mean. 1) Any $x \in k$ is algebraic over $k$ (it is the root of a linear polynomial). 2) $k[x]$ usually denotes the polynomial ring over $k$ in one variable $x$. But it seems you have a different interpretation since $x \in k$. $\endgroup$ – Paul Frost Sep 20 '18 at 15:33
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    $\begingroup$ Right! please check my update, hope it is clear $\endgroup$ – PerelMan Sep 20 '18 at 15:36
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    $\begingroup$ I added the "field-theory" and "extension'field" tags to your post. Cheers! $\endgroup$ – Robert Lewis Sep 20 '18 at 15:39

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 $x^{-1}\in k[x]$.


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.

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 $I:=\langle X^2-Y+1\rangle$.

  • $\begingroup$ Could you please explain why $k[x]\cong k[X]/\langle p \rangle$ ? is it the result of the universal property of the rupture field $ k[X]/\langle p \rangle$ which states that for any extension $k\subset L$ where $p$ has a root in $L$ there exists a unique $k$-morphism between $k[X]/\langle p \rangle$ and $L$ ? $\endgroup$ – PerelMan Sep 20 '18 at 16:33
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    $\begingroup$ @PerelMan Of course! Please, see the update of my answer $\endgroup$ – Jose Brox Sep 20 '18 at 17:16

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 $k(a) \subset k[a]$.


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

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


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

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 $y$:

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


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

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

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


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

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.

  • $\begingroup$ I think your implications are in reverse logical order: you say that since $x$ is algebraic, the dimension of $k[x]$ over $k$ is finite and then there is a linear dependence between powers of $x$. I don't see how you can prove this in this order (besides, the intermediate step is unnecessary), the straightforward one is to see that the powers are dependent because the element is algebraic, then conclude the finiteness of the dimension (if needed!) $\endgroup$ – Jose Brox Sep 21 '18 at 8:37

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