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Definition (algebraic independence over field extension)

[definition] Let $E/k$ be a field extension and $U \subseteq E$. We say that $U$ is algebraically dependent over $k$ if there exists $u_1, \ldots, u_n \in U$ and a nonzero polynomial $f\in k[x_1, \ldots, x_n]$ such that

We say that $B\subseteq E$ is algebraically independent (over $k$) if it is not algebraically dependent (over $k$).

We say that $E/k$ is purely transcendental if either $E = k$ or $E$ contains an algebraically independent subset $B$ and $E = k(B)$. [/definition]

Notice that this is basically a multivariate generalization of transcendental/algebraic elements over a field extension. In particular, recall that $\alpha$ is algebraic over $k$ if it is algebraically dependent over $k$ (is a root of a nonzero polynomial over $k$). If $\alpha$ isn’t algebraic over $k$, we said that $\alpha$ is transcendental over $k$ and also proved that $k(x) \cong k(\alpha)$ since there is an isomorphism $k[x] \to k[\alpha]$ that extends to quotient fields.

Example ($E = k(x)$)

[example] Suppose we let $E$ be the rational function field in a single variable over $k$. Then $k(x)/k$ is puerly transcendental with $B = \{x\}$. [/example]

Example ($E = \QQ(x_1, \ldots, x_n)$ and $k = \QQ$)

[example] This particular choice of a field $E$ arises commonly in cluster algebras (ambient field of a skew-symmetric) where we assume that $x_1, \ldots, x_n$ are algebraically independent. [/example]

Simple transcendental extension structure

[exercise=A-3.38(i)] Let $A$ and $B$ be isomorphic integral domains. If $\phi:A \to B$ is a ring isomorphism, then

is a ring isomorphism of the quotient fields $Q(A)$ and $Q(B)$. [/exercise]

[proof] (Well-definedness) Let $a/b = c/d$. Then $ad = bc$ and so $\phi(a)\phi(d) = \phi(b)\phi(c)$ and we get a well-defined map. Note also that $\phi$ has trivial kernel as well.

(Ring homomorphism) Obvious.

(Isomorphism) Obvious. [/proof]

[theorem] Let $\alpha$ be transcendental over $k$. Then $k[x] \cong k[\alpha]$. [/theorem]

[proof] Let $\phi:k[x] \to k[\alpha]$ be defined by $x \mapsto \alpha$. Surjectivity is immediate and injectivity is immediate by $\alpha$’s algebraic independence over $k$. Clearly, $\phi$ is also a ring homomorphism so the previous result tells us that $\phi$ lifts to a isomorphism of $k[x]$ and $k[\alpha]$ (notice that $k[\alpha]$ is actually an integral domain by transcendence of $\alpha$). [/proof]

Lemma B-2.45 (Purely transcendental extension $k(B)/k$ structure with $B$ finite)

[theorem=A-3.25] Let $\phi:R \to S$ be a homomorphism of commutative rings. If $s_1, \ldots, s_n \in S$, then there exists a unique homomorphism

with $\Phi(x_i) = s_i$ for all $i$ and $\Phi(r) = \phi(r)$ for all $r\in R$. [/theorem]

[proof] We proceed by induction on $n\ge 1$.

(Base case: $n = 1$) Obvious.

(Inductive step: $n > 1$) Follows immediately from the base case and inductive hypothesis by recalling

[/proof]

[lemma=B-2.45] Let $E/k$ be a purely transcendental extension with $E = k(B)$ where $B = \{u_1, \ldots, u_n\}$ be a finite algebraically independent subset. If $k(x_1, \ldots, x_n)$ is the function field with indeterminates $x_1, \ldots, x_n$, then there is an isomorphism

[/lemma]

[proof] By the theorem above, there is a unique ring homomorphism

that sends $x_i$ to $u_i$ for all $i$ and fixes the base field $k$. By the same argumentation for the simple case we gave in the previous section, $\phi$ extends to the desired map. [/proof]

Proposition B-2.47 (Purely transcendental extension $k(B)/k$ in general)

[proposition] Let $E/k$ be a purely transcendental extension where $E = k(B)$ and $B\subseteq E$ is an algebraically independent subset. Then $E\cong k(X)$ as rings with $|X| = |B|$. [/proposition]

Definition (Transcendence basis)

[definition] If $E/k$ is an extension field, then a transcendence basis is a maximal algebraically independent subset of $E$ over $k$. [/definition]

Theorem B-2.51 (Every field extension has a transcendence basis)

[theorem] If $E/k$ is an exteension field, then $E$ has a transcendence basis. In fact, every algebraically independent subset is part of a transcendence basis. [/theorem]

Theorem B-2.52

[theorem] If $B$ is a transcendence basis, then $k(B)/k$ is purely transcendental and $E/k(B)$ is algebraic. [/theorem]

Theorem B-2.53 (Transcendence degree is well-defined)

[theorem] If $B$ and $C$ are transcendence bases of an extension field $E/k$, then $|B| = |C|$. [/theorem]

Definition (Transcendence degree)

[definition] The transcendence degree of an extension field $E/k$ is defined by

where $|B|$ is a transcendence basis of $E/k$. [/definition]