Algebraic Extensions

Field Extensions

We saw by §Fields, ⁋Proposition 2 that a morphism between fields is either injective or the zero map. In this post we examine the former case.

We call an injective field morphism a field extension. Then, for a fixed field \(\mathbb{K}\in\Field\), the under category of \(\mathbb{K}\) becomes the category of extensions of \(\mathbb{K}\).

Although this notation differs slightly from that of [Category Theory] §Category, ⁋Example 13, we often denote a field extension \(\mathbb{K}\rightarrow \mathbb{L}\) by \(\mathbb{L}/\mathbb{K}\). Then, whenever a field extension \(\mathbb{L}/\mathbb{K}\) is given, via the injective map \(\mathbb{K}\hookrightarrow\mathbb{L}\) we can identify \(\mathbb{K}\) with a subfield of \(\mathbb{L}\). However, if \(\mathbb{L}=\mathbb{K}\) and \(\mathbb{K}\hookrightarrow\mathbb{L}=\mathbb{K}\) is an endomorphism, such an identification may cause confusion, so in this case we do not identify \(\mathbb{K}\) with a subfield of \(\mathbb{L}\).

By definition, given two extensions \(\mathbb{K} \rightarrow \mathbb{L}_1\) and \(\mathbb{K} \rightarrow \mathbb{L}_2\), the following commutative diagram

is a morphism between them. Here, since both \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are fields, the morphism \(\mathbb{L}_1 \rightarrow \mathbb{L}_2\) must be injective. Subject to the caution above, in this case we call \(\mathbb{L}_1\) a subextension of \(\mathbb{L}_2\).

Thus any field extension \(\mathbb{L}/\mathbb{K}\) can be regarded as an associative unital \(\mathbb{K}\)-algebra (which is itself a field).

Remark We consider \(\mathbb{K}\)-algebras and homomorphisms between them in order to treat situations similar to the above; henceforth in our posts the category \(\Alg{\mathbb{K}}\) will always be the category of unital associative \(\mathbb{K}\)-algebras. That is, by a \(\mathbb{K}\)-algebra we shall always mean a unital associative \(\mathbb{K}\)-algebra, and by a \(\mathbb{K}\)-algebra homomorphism we shall always mean a unital \(\mathbb{K}\)-algebra homomorphism.

Any \(\mathbb{K}\)-algebra is also a \(\mathbb{K}\)-module, so its dimension is well-defined. ([Multilinear Algebra] §Bases, ⁋Proposition 6)

Definition 1 For any \(\mathbb{K}\)-algebra \(A\), we call \(\dim_{\mathbb{K}}A\) the degree of \(A\) and denote it by \([A:\mathbb{K}]\).

Then the following is obvious from the definition.

Proposition 2 For a field extension \(\mathbb{L}_2/\mathbb{L}_1/\mathbb{K}\), \([\mathbb{L}_2:\mathbb{K}]=[\mathbb{L}_2:\mathbb{L}_1][\mathbb{L}_1:\mathbb{K}]\) holds.

More generally, for a field \(\mathbb{K}\) and any \(\mathbb{K}\)-algebra \(E\), we can define \([E:\mathbb{K}]\) as \(\dim_\mathbb{K}E\). Then the following proposition is simple linear algebra.

Proposition 3 For a finite degree \(\mathbb{K}\)-algebra \(E\), if \(x\in E\) is a non-zerodivisor in \(E\), then \(x\) is an invertible element of \(E\).

Proof

By assumption, \(E\) is a finite-dimensional \(\mathbb{K}\)-algebra, and since \(x\) is a non-zerodivisor of \(E\), the function

\[E \rightarrow E;\qquad y\mapsto xy\]

is injective. Now since \(E\) is finite-dimensional, injectivity of the above linear map is equivalent to surjectivity, and therefore \(xy=1\) holds for some \(y\in E\). Here, if we define \(L_x, L_y\) by \(z\mapsto xz\) and \(z\mapsto yz\) respectively, then \(L_x\) is bijective and \(L_x\circ L_y=\id\), so \(L_y=L_x^{-1}\); thus evaluating \(L_y\circ L_x=\id\) at \(z=1\) yields \(yx=1\), from which we obtain the desired result.

In particular, if a finite-dimensional \(\mathbb{K}\)-algebra \(E\) is an integral domain, then \(E\) is necessarily a field.

Meanwhile, for any ring \(A\), we have denoted the polynomial ring with coefficients in \(A\) in the variable \(\x\) by \(A[\x]\), which can be thought of as the smallest algebra containing \(A\) and the variable \(\x\). In a similar manner, to adjoin to a field \(\mathbb{K}\) an element \(\x\), this time we will also have to add the inverses.

Definition 4 For a field extension \(\mathbb{L}/\mathbb{K}\) and a subset \(A\subseteq \mathbb{L}\), we denote the smallest subextension containing \(A\) in \(\mathbb{L}\) by \(\mathbb{K}(A)\).

In this definition, \(\mathbb{L}\) is needed only to define \(A\), and regardless of what \(\mathbb{L}\) actually is, \(\mathbb{K}(A)\) will be an isomorphic field. For this reason, we often fix for \(\mathbb{K}\) a (very large) field extension \(\Omega\) (without worrying about what \(\Omega\) is) and consider subsets \(M,N\) of this extension.

Proposition 5 For a suitable extension of \(\mathbb{K}\) and two subsets \(M,N\) of it, the following formula

\[\mathbb{K}(M \cup N) = \mathbb{K}(M)(N) = \mathbb{K}(N)(M)\]

holds.

The proof of this is almost obvious from the minimality of the definition.

Meanwhile, to obtain the field \(\mathbb{K}(A)\) of Definition 4, one can fix a \(\mathbb{K}\)-extension \(\mathbb{L}\) and then take the intersection of all subextensions containing \(A\) in \(\mathbb{L}\). On the other hand, since we have shown that morphisms in the category of extensions of \(\mathbb{K}\) are only extensions, the following holds.

Proposition 6 Let \(\mathcal{F}\) be a set of subfields of a field \(E\); with the inclusion relation \(\subseteq\), it becomes a directed set. In particular, the union \(L\) of the fields belonging to \(\mathcal{F}\) is a field.

If \(\mathbb{L}=\mathbb{K}(A)\) for some finite set \(A\), we call the extension \(\mathbb{L}/\mathbb{K}\) a finite extension. Then in particular a finite degree field extension is a finite extension, because a basis of \(\mathbb{L}\) as a \(\mathbb{K}\)-vector space will serve as a set of generators of \(\mathbb{L}\) as a field.

Now let two \(\mathbb{K}\)-extensions \(\mathbb{L}_1/\mathbb{K}\), \(\mathbb{L}_2/\mathbb{K}\) be given. Then we can consider the smallest extension containing both \(\mathbb{L}_1\) and \(\mathbb{L}_2\).

Definition 7 For two \(\mathbb{K}\)-extensions \(\mathbb{L}_1/\mathbb{K}\), \(\mathbb{L}_2/\mathbb{K}\), that a \(\mathbb{K}\)-extension \(\mathbb{K} \rightarrow \mathbb{M}\) is their composite means that, for the following diagram,

there exist \(\mathbb{K}\)-algebra homomorphisms \(u_1:\mathbb{L}_1 \rightarrow \mathbb{M}\) and \(u_2:\mathbb{L}_2 \rightarrow \mathbb{M}\) making it commute, and \(\mathbb{M}=\mathbb{K}(u_1(\mathbb{L}_1)\cup u_2(\mathbb{L}_2))\) holds.

Concretely, this can be written as follows.

Proposition 8 Let two \(\mathbb{K}\)-extensions \(\mathbb{L}_1, \mathbb{L}_2\) be given.

1. For their composite field \(\mathbb{M}\) and extensions \(u_i: \mathbb{L}_i \rightarrow \mathbb{M}\), the following formula

   \[\mathbb{L}_1\otimes_\mathbb{K} \mathbb{L}_2 \rightarrow \mathbb{M};\qquad x_1\otimes x_2\mapsto u_1(x_1)u_2(x_2)\]

   defines a function \(u_1\ast u_2: \mathbb{L}_1\otimes_\mathbb{K} \mathbb{L}_2 \rightarrow \mathbb{M}\) whose kernel \(\ker (u_1\ast u_2)\) is a prime ideal.

2. Conversely, in \(\mathbb{L}_1\otimes_\mathbb{K} \mathbb{L}_2\), for any prime ideal \(\mathfrak{p}\), there exist a suitable composite field \(\mathbb{M}\) and extensions \(u_i: \mathbb{L}_i \rightarrow \mathbb{M}\) such that \(\mathfrak{p}\) is the kernel of \(u_1\ast u_2\).

Proof

1. For \(u_1\ast u_2\), its image \(\im(u_1\ast u_2)\) is a subring of the field \(\mathbb{M}\), and hence an integral domain. Now the given claim is obvious from [Algebraic Structures] §Field of Fractions, ⁋Proposition 9 and [Algebraic Structures] §Quotient Rings and Ring Isomorphisms, ⁋Theorem 3.

2. Conversely, let \(\mathfrak{p}\) be a prime ideal of \(\mathbb{L}_1\otimes_\mathbb{K}\mathbb{L}_2\), and let the field of fractions of the integral domain \((\mathbb{L}_1\otimes_\mathbb{K}\mathbb{L}_2)/\mathfrak{p}\) be \(\mathbb{M}=\Frac((\mathbb{L}_1\otimes_\mathbb{K}\mathbb{L}_2)/\mathfrak{p})\). Then for each \(x_1\in \mathbb{L}_1\) and \(x_2\in \mathbb{L}_2\), defining \(u_1(x_1)\) to be the image of \(x_1\otimes 1\) in \(\mathbb{M}\) and \(u_2(x_2)\) to be the image of \(1\otimes x_2\) in \(\mathbb{M}\), we see that these satisfy the required conditions.

Moreover, it is also obvious that the composite field obtained from the second result is uniquely determined up to isomorphism. Meanwhile, for any two \(\mathbb{K}\)-extensions \(\mathbb{L}_1, \mathbb{L}_2\), since \(\mathbb{L}_1\otimes_\mathbb{K} \mathbb{L}_2\) always has a prime ideal ([Algebraic Structures] §Definition of a Ring, ⁋Theorem 10 (Krull)), we can verify that any two \(\mathbb{K}\)-extensions have a composite field.

Algebraic Extensions

For \(\mathbb{K}\), let us fix a suitable extension \(\Omega\). Unless stated otherwise, all field extensions are assumed to be subextensions of \(\Omega\).

In \(\Omega\), for any two \(\mathbb{K}\)-subalgebras \(E,F\), considering the multiplication map \(\mu: E\otimes_\mathbb{K}F \rightarrow \Omega\), its image \(G\) is the subring generated by \(E\cup F\) in \(\Omega\).

Definition 9 In the above situation, if the multiplication map \(\mu: E\otimes_\mathbb{K}F \rightarrow G\) is an isomorphism, we say that \(E\) and \(F\) are linearly disjoint.

It is not difficult to see that for \(E\) and \(F\), given two \(\mathbb{K}\)-bases \((x_i)_{i\in I}\) and \((y_j)_{j\in J}\), this is equivalent to \((x_iy_j)_{i\in I,j\in J}\) being linearly independent.

In particular, when \(E,F\) are \(\mathbb{K}\)-extensions, we obtain the following proposition.

Proposition 10 For two \(\mathbb{K}\)-extensions \(\mathbb{L}_1, \mathbb{L}_2\), the following hold.

1. If \(\mathbb{L}_2\) has finite degree, the subring generated by \(\mathbb{L}_1 \cup \mathbb{L}_2\) in \(\Omega\) is a field, which coincides with \(\mathbb{L}_1(\mathbb{L}_2)\). Moreover, the degree of \(\mathbb{L}_1(\mathbb{L}_2)\) over \(\mathbb{L}_1\) is also finite, and

   \[[\mathbb{L}_1(\mathbb{L}_2) : \mathbb{L}_1] \leq [\mathbb{L}_2 : \mathbb{K}]\]

   with equality holding when \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint. In this case, \(\mathbb{L}_1(\mathbb{L}_2)\) and \(\mathbb{L}_1 \otimes_\mathbb{K} \mathbb{L}_2\) are \(\mathbb{L}_1\)-isomorphic.

2. In addition to the above conditions, assume further that the degree of \(\mathbb{L}_1\) is also finite. Then the degree of \(\mathbb{L}_1(\mathbb{L}_2) = \mathbb{K}(\mathbb{L}_1 \cup \mathbb{L}_2)\) is also finite, and

   \[[\mathbb{K}(\mathbb{L}_1 \cup \mathbb{L}_2) : \mathbb{K}] \leq [\mathbb{L}_1 : \mathbb{K}][\mathbb{L}_2 : \mathbb{K}]\]

   with equality holding when \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint.

Proof

1. Let the subring generated by \(\mathbb{L}_1 \cup \mathbb{L}_2\) in \(\Omega\) be \(G\). If \((y_j)_{1 \leq j \leq n}\) is a basis of \(\mathbb{L}_2\) over \(\mathbb{K}\), then \(G\) is generated as an \(\mathbb{L}_1\)-vector space by the \(y_j\). Then \(G\) has finite rank \(\leq n\) as an \(\mathbb{L}_1\)-algebra. Now, since \(G\) is contained in the field \(\Omega\), it is an integral domain, and therefore is a field by Proposition 3. Consequently, \(G=\mathbb{L}_1(\mathbb{L}_2)\), and

   \[[\mathbb{L}_1(\mathbb{L}_2) : \mathbb{L}_1] \leq [\mathbb{L}_2 : \mathbb{K}]\]

   holds.
   Moreover, if \([\mathbb{L}_1(\mathbb{L}_2) : \mathbb{L}_1] = [\mathbb{L}_2 : \mathbb{K}]\), then the \(y_j\) must be linearly independent over \(\mathbb{L}_1\); that is, \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint.

2. This is obvious from the following formula:

   \[[\mathbb{L}_1(\mathbb{L}_2) : \mathbb{K}] = [\mathbb{L}_1(\mathbb{L}_2) : \mathbb{L}_1][\mathbb{L}_1 : \mathbb{K}]\]

In general, for two \(\mathbb{K}\)-extensions \(\mathbb{L}_1,\mathbb{L}_2\), we mentioned earlier that the ring \(\mathbb{K}[\mathbb{L}_1\cup \mathbb{L}_2]\) is not a field. However, the formula

\[\Frac(\mathbb{K}[\mathbb{L}_1\cup\mathbb{L}_2])=\mathbb{K}(\mathbb{L}_1\cup\mathbb{L}_2)\]

holds. More generally, let \(S_i\subseteq \mathbb{L}_i\) be subsets satisfying \(\Frac(\mathbb{K}[S_i])=\mathbb{L}_i\). If \(G\) is the ring generated by \(\mathbb{K}\cup S_1\cup S_2\), we obtain the isomorphism

\[\Frac(G)\cong \mathbb{K}(\mathbb{L}_1\cup\mathbb{L}_2)\]

The following proposition extends this observation to the language of linearly disjoint extensions.

Proposition 11 Let \(E_1, E_2\) in \(\Omega\) be \(\mathbb{K}\)-subalgebras. If \(\mathbb{L}_i=\Frac(E_i)\), then \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint if and only if \(E_1\) and \(E_2\) are linearly disjoint.

Proof

One direction is obvious, so assume that \(E_1, E_2\) are linearly disjoint. Then we can first show that \(E_1\) and \(\mathbb{L}_2\) are linearly disjoint, because any family in \(\Omega\) that is \(E_2\)-free is also \(\mathbb{L}_2\)-free. Now, by the same logic, \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint.

Meanwhile, since a linear combination of an arbitrary family consists only of finite sums (even if the family is infinite), the following holds.

Proposition 12 For a field \(\mathbb{K}\), consider two extensions \(\mathbb{L}_1\), \(\mathbb{L}_2\). If \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are linearly disjoint, then every subextension of \(\mathbb{L}_1\) and every subextension of \(\mathbb{L}_2\) are also linearly disjoint over \(\mathbb{K}\). Conversely, if for the \(\mathbb{L}_i\), all finitely generated subextensions \(\mathbb{L}_i'\) are such that \(\mathbb{L}_1'\) and \(\mathbb{L}_2'\) are linearly disjoint, then \(\mathbb{L}_1\) and \(\mathbb{L}_2\) are also linearly disjoint.

In other words, whether two arbitrary extensions are linearly disjoint can be checked just by looking at arbitrary finite subextensions of the two extensions.

Proposition 13 Let three \(\mathbb{K}\)-extensions \(\mathbb{L},\mathbb{M}_1,\mathbb{M}_2\) be given, and suppose \(\mathbb{M}_1 \subseteq \mathbb{M}_2\). Then \(\mathbb{L}\) and \(\mathbb{M}_2\) being linearly disjoint is equivalent to \(\mathbb{L}\) and \(\mathbb{M}_1\) being linearly disjoint and at the same time \(\mathbb{L}(\mathbb{M}_1)\) and \(\mathbb{M}_2\) being linearly disjoint over \(\mathbb{M}_1\).

Proof

First assume that \(\mathbb{L}\) and \(\mathbb{M}_2\) are linearly disjoint. Then by Proposition 12, \(\mathbb{L}\) and \(\mathbb{M}_1\) are also linearly disjoint. On the other hand, a basis of \(\mathbb{L}\) over \(\mathbb{K}\) is also a basis of \(\mathbb{M}_1[\mathbb{L}]\) over \(\mathbb{M}_1\). But by assumption this basis is \(\mathbb{M}_2\)-free, so \(\mathbb{M}_1[\mathbb{L}]\) and \(\mathbb{M}_2\) are linearly disjoint over \(\mathbb{M}_1\). Also, by Proposition 11, \(\mathbb{L}(\mathbb{M}_1) = \mathbb{M}_1(\mathbb{L})\) and \(\mathbb{M}_2\) are also linearly disjoint over \(\mathbb{M}_1\).

Now let us show the converse. As above, considering a basis of \(\mathbb{L}\) over \(\mathbb{K}\), denoted \(B\), from the hypothesis \(B\) is \(\mathbb{M}_1\)-free. Therefore \(B\) is a basis of \(\mathbb{M}_1[\mathbb{L}]\) over \(\mathbb{M}_1\), and again by hypothesis \(\mathbb{M}_1[\mathbb{L}]\) and \(\mathbb{M}_2\) are linearly disjoint over \(\mathbb{M}_1\), so we obtain the desired result.

Consider a field \(\mathbb{K}\) and a \(\mathbb{K}\)-algebra \(E\). Then for any \(x\in E\), exactly one of the following two holds.

1. \((x^n)_{n\geq 0}\) is \(\mathbb{K}\)-free.
2. \(1,x,\cdots, x^{n-1}\) are \(\mathbb{K}\)-linearly dependent for some \(n\).

Definition 14 In the above situation, if the first case holds we call \(x\in E\) transcendental, and if the second case holds we call \(x\) algebraic.

Now suppose \(x\in E\) is algebraic. Then, for \(1,x,\ldots, x^{n}\) to be \(\mathbb{K}\)-linearly dependent, the smallest such \(n\) is called the degree of \(x\), and for the linear combination at this time

\[a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0\]

the following polynomial

\[f(\x)=\x^n+\sum_{k=0}^{n-1}\frac{a_k}{a_n}\x^k\]

is called the minimal polynomial of \(x\).

Then the following theorem holds, and its proof is not very difficult either.

Theorem 15 For a \(\mathbb{K}\)-algebra \(E\) and an algebraic element \(x\in E\), let the degree of \(x\) be \(n\) and its minimal polynomial be \(f\). Then the following hold.

1. For \(g \in \mathbb{K}[\x]\), \(g(x) = 0\) if and only if \(g\) is a multiple of \(f\).
2. Define \(\mathbb{K}[\x] \rightarrow \mathbb{K}[x]\) by \(g\mapsto g(x)\). Then this morphism factors through the quotient algebra \(\mathbb{K}[\x]/(f)\), and the resulting map \(\mathbb{K}[\x]/(f) \rightarrow \mathbb{K}[x]\) is an isomorphism. Moreover, in this case \(1, x, \dots, x^{n-1}\) form a basis of \(\mathbb{K}[x]\) over \(\mathbb{K}\), and therefore \([\mathbb{K}[x] : \mathbb{K}] = n\) holds.
3. If \(E\) is an integral domain, then \(\mathbb{K}[x]\) is a field, and \(f\in \mathbb{K}[\x]\) is the unique monic irreducible polynomial satisfying \(f(x) = 0\).
4. \(x\) is an invertible element in \(E\) if and only if \(f(0) \neq 0\), in which case \(x^{-1} \in \mathbb{K}[x]\).

Also, for an extension \(\mathbb{K}\hookrightarrow \mathbb{L}\) and an \(\mathbb{L}\)-algebra \(E\), we know that if an element \(x\in E\) is algebraic over \(\mathbb{K}\), then \(x\) is also algebraic over \(\mathbb{L}\) and its degree does not exceed its degree over \(\mathbb{K}\).

Definition 16 A field extension \(\mathbb{L}/\mathbb{K}\) in which every element is algebraic is called an algebraic extension. A field extension that is not algebraic is called a transcendental extension.

Then it is not difficult to see that an extension \(\mathbb{L}/\mathbb{K}\) being algebraic is equivalent to every subalgebra of \(\mathbb{L}\) over \(\mathbb{K}\) being a field. Also, the following proposition is obvious.

Proposition 17 A degree \(n\) \(\mathbb{K}\)-extension \(\mathbb{L}\) is necessarily an algebraic extension, and the degree of any element of \(\mathbb{L}\) is a divisor of \(n\).

Proof

\[[\mathbb{L}:\mathbb{K}]=[\mathbb{L}:\mathbb{K}(x)][\mathbb{K}(x):\mathbb{K}].\]

Extending this inductively, we obtain the following.

Theorem 18 Suppose that a finitely generated \(\mathbb{K}\)-extension \(\mathbb{L}\) is generated by algebraic elements \(a_1, \dots, a_m\). Then \(\mathbb{L}/\mathbb{K}\) is a finite degree extension. Moreover, if for each \(i\), we let the degree of \(a_i\) over \(\mathbb{K}(a_1, \dots, a_{i-1})\) be \(n_i\), then the degree of \(\mathbb{L}\) over \(\mathbb{K}\) is \(n_1 n_2 \cdots n_m\), and the following elements

\[a_1^{\nu_1} a_2^{\nu_2} \cdots a_m^{\nu_m}\qquad (0 \leq \nu_i \leq n_i - 1)\]

form a basis of \(\mathbb{L}\) over \(\mathbb{K}\).

In particular, for a set \(A\) consisting only of algebraic elements, \(\mathbb{K}(A)=\mathbb{K}[A]\) holds. Moreover, algebraic extensions are transitive. That is, the following proposition holds.

Proposition 19 For a field extension \(\mathbb{M}/\mathbb{L}/\mathbb{K}\), \(\mathbb{M}\) is algebraic over \(\mathbb{K}\) if and only if \(\mathbb{L}\) is algebraic over \(\mathbb{K}\) and at the same time \(\mathbb{M}\) is algebraic over \(\mathbb{L}\).

Proof

One direction is obvious, so it suffices to show the converse. Assume that \(\mathbb{L}\) is algebraic over \(\mathbb{K}\) and at the same time \(\mathbb{M}\) is algebraic over \(\mathbb{L}\), and choose from \(\mathbb{M}\) an arbitrary element \(x\). We must show that \(x\) is algebraic over \(\mathbb{K}\).

First, by assumption \(x\) is algebraic over \(\mathbb{L}\). Let \(g \in \mathbb{L}[\x]\) be the minimal polynomial of \(x\), and let the set of coefficients of \(g\) be \(A\). Then \(g \in \mathbb{K}(A)[\x]\), and therefore \(x\) is algebraic over \(\mathbb{K}(A)\).

Also, \(\mathbb{K}(A \cup \{x\}) = \mathbb{K}(A)(x)\) has finite degree over \(\mathbb{K}(A)\). \(A\) is a finite set consisting of the coefficients of \(g\), and since \(A \subseteq \mathbb{L}\) and \(\mathbb{L}\) is algebraic over \(\mathbb{K}\), by Theorem 18, \(\mathbb{K}(A)\) has finite degree over \(\mathbb{K}\). From this, \(\mathbb{K}(A \cup \{x\})\) has finite degree over \(\mathbb{K}\), and therefore \(x\) is algebraic over \(\mathbb{K}\).

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References

[Bou] N. Bourbaki. Algebra II: Chapters 4–7. Springer, 2003.