Definition 9.8.1. Consider a field extension $F/E$. An element $\alpha \in F$ is said to be *algebraic* over $E$ if $\alpha $ is the root of some nonzero polynomial with coefficients in $E$. If all elements of $F$ are algebraic then $F$ is said to be an *algebraic extension* of $E$.

## 9.8 Algebraic extensions

An important class of extensions are those where every element generates a finite extension.

By Lemma 9.6.8, the subextension $E(\alpha )$ is isomorphic either to the rational function field $E(t)$ or to a quotient ring $E[t]/(P)$ for $P \in E[t]$ an irreducible polynomial. In the latter case, $\alpha $ is algebraic over $E$ (in fact, the proof of Lemma 9.6.8 shows that we can pick $P$ such that $\alpha $ is a root of $P$); in the former case, it is not.

Example 9.8.2. The field $\mathbf{C}$ is algebraic over $\mathbf{R}$. Namely, if $\alpha = a + ib$ in $\mathbf{C}$, then $\alpha ^2 - 2a\alpha + a^2 + b^2 = 0$ is a polynomial equation for $\alpha $ over $\mathbf{R}$.

Example 9.8.3. Let $X$ be a compact Riemann surface, and let $f \in \mathbf{C}(X) - \mathbf{C}$ any nonconstant meromorphic function on $X$ (see Example 9.3.6). Then it is known that $\mathbf{C}(X)$ is algebraic over the subextension $\mathbf{C}(f)$ generated by $f$. We shall not prove this.

Lemma 9.8.4. Let $K/E/F$ be a tower of field extensions.

If $\alpha \in K$ is algebraic over $F$, then $\alpha $ is algebraic over $E$.

If $K$ is algebraic over $F$, then $K$ is algebraic over $E$.

**Proof.**
This is immediate from the definitions.
$\square$

We now show that there is a deep connection between finiteness and being algebraic.

Lemma 9.8.5. A finite extension is algebraic. In fact, an extension $E/k$ is algebraic if and only if every subextension $k(\alpha )/k$ generated by some $\alpha \in E$ is finite.

In general, it is very false that an algebraic extension is finite.

**Proof.**
Let $E/k$ be finite, say of degree $n$. Choose $\alpha \in E$. Then the elements $\{ 1, \alpha , \ldots , \alpha ^ n\} $ are linearly dependent over $E$, or we would necessarily have $[E : k] > n$. A relation of linear dependence now gives the desired polynomial that $\alpha $ must satisfy.

For the last assertion, note that a monogenic extension $k(\alpha )/k$ is finite if and only if $\alpha $ is algebraic over $k$, by Examples 9.7.4 and 9.7.6. So if $E/k$ is algebraic, then each $k(\alpha )/k$, $\alpha \in E$, is a finite extension, and conversely. $\square$

We can extract a lemma of the last proof (really of Examples 9.7.4 and 9.7.6): a monogenic extension is finite if and only if it is algebraic. We shall use this observation in the next result.

Lemma 9.8.6. Let $k$ be a field, and let $\alpha _1, \alpha _2, \ldots , \alpha _ n$ be elements of some extension field such that each $\alpha _ i$ is algebraic over $k$. Then the extension $k(\alpha _1, \ldots , \alpha _ n)/k$ is finite. That is, a finitely generated algebraic extension is finite.

**Proof.**
Indeed, each extension $k(\alpha _{1}, \ldots , \alpha _{i+1})/k(\alpha _1, \ldots , \alpha _{i})$ is generated by one element and algebraic, hence finite. By multiplicativity of degree (Lemma 9.7.7) we obtain the result.
$\square$

The set of complex numbers that are algebraic over $\mathbf{Q}$ are simply called the *algebraic numbers.* For instance, $\sqrt{2}$ is algebraic, $i$ is algebraic, but $\pi $ is not. It is a basic fact that the algebraic numbers form a field, although it is not obvious how to prove this from the definition that a number is algebraic precisely when it satisfies a nonzero polynomial equation with rational coefficients (e.g. by polynomial equations).

Lemma 9.8.7. Let $E/k$ be a field extension. Then the elements of $E$ algebraic over $k$ form a subextension of $E/k$.

**Proof.**
Let $\alpha , \beta \in E$ be algebraic over $k$. Then $k(\alpha , \beta )/k$ is a finite extension by Lemma 9.8.6. It follows that $k(\alpha + \beta ) \subset k(\alpha , \beta )$ is a finite extension, which implies that $\alpha + \beta $ is algebraic by Lemma 9.8.5. Similarly for the difference, product and quotient of $\alpha $ and $\beta $.
$\square$

Many nice properties of field extensions, like those of rings, will have the property that they will be preserved by towers and composita.

Lemma 9.8.8. Let $E/k$ and $F/E$ be algebraic extensions of fields. Then $F/k$ is an algebraic extension of fields.

**Proof.**
Choose $\alpha \in F$. Then $\alpha $ is algebraic over $E$. The key observation is that $\alpha $ is algebraic over a finitely generated subextension of $k$. That is, there is a finite set $S \subset E$ such that $\alpha $ is algebraic over $k(S)$: this is clear because being algebraic means that a certain polynomial in $E[x]$ that $\alpha $ satisfies exists, and as $S$ we can take the coefficients of this polynomial. It follows that $\alpha $ is algebraic over $k(S)$. In particular, the extension $k(S, \alpha )/ k(S)$ is finite. Since $S$ is a finite set, and $k(S)/k$ is algebraic, Lemma 9.8.6 shows that $k(S)/k$ is finite. Using multiplicativity (Lemma 9.7.7) we find that $k(S,\alpha )/k$ is finite, so $\alpha $ is algebraic over $k$.
$\square$

The method of proof in the previous argument — that being algebraic over $E$ was a property that *descended* to a finitely generated subextension of $E$ — is an idea that recurs throughout algebra. It often allows one to reduce general commutative algebra questions to the Noetherian case for example.

Lemma 9.8.9. Let $E/F$ be an algebraic extension of fields. Then the cardinality $|E|$ of $E$ is at most $\max (\aleph _0, |F|)$.

**Proof.**
Let $S$ be the set of nonconstant polynomials with coefficients in $F$. For every $P \in S$ the set of roots $r(P, E) = \{ \alpha \in E \mid P(\alpha ) = 0\} $ is finite (details omitted). Moreover, the fact that $E$ is algebraic over $F$ implies that $E = \bigcup _{P \in S} r(P, E)$. It is clear that $S$ has cardinality bounded by $\max (\aleph _0, |F|)$ because the cardinality of a countable product of copies of $F$ has cardinality at most $\max (\aleph _0, |F|)$. Thus so does $E$.
$\square$

Lemma 9.8.10. Let $E/F$ be a finite or more generally an algebraic extension of fields. Any subring $F \subset R \subset E$ is a field.

**Proof.**
Let $\alpha \in R$ be nonzero. Then $1, \alpha , \alpha ^2, \ldots $ are contained in $R$. By Lemma 9.8.5 we find a nontrivial relation $a_0 + a_1 \alpha + \ldots + a_ d \alpha ^ d = 0$. We may assume $a_0 \not= 0$ because if not we can divide the relation by $\alpha $ to decrease $d$. Then we see that

which proves that the inverse of $\alpha $ is the element $a_0^{-1} (- a_1 - \ldots - a_ d \alpha ^{d - 1})$ of $R$. $\square$

Lemma 9.8.11. Let $E/F$ an algebraic extension of fields. Any $F$-algebra map $f : E \to E$ is an automorphism.

**Proof.**
If $E/F$ is finite, then $f : E \to E$ is an $F$-linear injective map (Lemma 9.6.1) of finite dimensional vector spaces, and hence bijective. In general we still see that $f$ is injective. Let $\alpha \in E$ and let $P \in F[x]$ be a polynomial such that $P(\alpha ) = 0$. Let $E' \subset E$ be the subfield of $E$ generated by the roots $\alpha = \alpha _1, \ldots , \alpha _ n$ of $P$ in $E$. Then $E'$ is finite over $F$ by Lemma 9.8.6. Since $f$ preserves the set of roots, we find that $f|_{E'} : E' \to E'$. Hence $f|_{E'}$ is an isomorphism by the first part of the proof and we conclude that $\alpha $ is in the image of $f$.
$\square$

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Comment #8463 by Abhishek Jha on