The Stacks project

Lemma 9.26.3. Let $E/F$ be a field extension. A transcendence basis of $E$ over $F$ exists. Any two transcendence bases have the same cardinality.

Proof. Let $A$ be an algebraically independent subset of $E$. Let $G$ be a subset of $E$ containing $A$ that generates $E/F$. We claim we can find a transcendence basis $B$ such that $A \subset B \subset G$. To prove this, consider the collection $\mathcal{B}$ of algebraically independent subsets whose members are subsets of $G$ that contain $A$. Define a partial ordering on $\mathcal{B}$ using inclusion. Then $\mathcal{B}$ contains at least one element $A$. The union of the elements of a totally ordered subset $T$ of $\mathcal{B}$ is an algebraically independent subset of $E$ over $F$ since any algebraic dependence relation would have occurred in one of the elements of $T$ (since polynomials only involve finitely many variables). The union also contains $A$ and is contained in $G$. By Zorn's lemma, there is a maximal element $B \in \mathcal{B}$. Now we claim $E$ is algebraic over $F(B)$. This is because if it wasn't then there would be an element $f \in G$ transcendental over $F(B)$ since $F(G) = E$. Then $B \cup \{ f\} $ would be algebraically independent contradicting the maximality of $B$. Thus $B$ is our transcendence basis.

Let $B$ and $B'$ be two transcendence bases. Without loss of generality, we can assume that $|B'| \leq |B|$. Now we divide the proof into two cases: the first case is that $B$ is an infinite set. Then for each $\alpha \in B'$, there is a finite set $B_{\alpha } \subset B$ such that $\alpha $ is algebraic over $F(B_{\alpha })$ since any algebraic dependence relation only uses finitely many indeterminates. Then we define $B^* = \bigcup _{\alpha \in B'} B_{\alpha }$. By construction, $B^* \subset B$, but we claim that in fact the two sets are equal. To see this, suppose that they are not equal, say there is an element $\beta \in B \setminus B^*$. We know $\beta $ is algebraic over $F(B')$ which is algebraic over $F(B^*)$. Therefore $\beta $ is algebraic over $F(B^*)$, a contradiction. So $|B| \leq |\bigcup _{\alpha \in B'} B_{\alpha }|$. Now if $B'$ is finite, then so is $B$ so we can assume $B'$ is infinite; this means

\[ |B| \leq |\bigcup \nolimits _{\alpha \in B'} B_{\alpha }| = |B'| \]

because each $B_\alpha $ is finite and $B'$ is infinite. Therefore in the infinite case, $|B| = |B'|$.

Now we need to look at the case where $B$ is finite. In this case, $B'$ is also finite, so suppose $B = \{ \alpha _1, \ldots , \alpha _ n\} $ and $B' = \{ \beta _1, \ldots , \beta _ m\} $ with $m \leq n$. We perform induction on $m$: if $m = 0$ then $E/F$ is algebraic so $B = \emptyset $ so $n = 0$. If $m > 0$, there is an irreducible polynomial $f \in F[x, y_1, \ldots , y_ n]$ such that $f(\beta _1, \alpha _1, \ldots , \alpha _ n) = 0$ and such that $x$ occurs in $f$. Since $\beta _1$ is not algebraic over $F$, $f$ must involve some $y_ i$ so without loss of generality, assume $f$ uses $y_1$. Let $B^* = \{ \beta _1, \alpha _2, \ldots , \alpha _ n\} $. We claim that $B^*$ is a basis for $E/F$. To prove this claim, we see that we have a tower of algebraic extensions

\[ E/ F(B^*, \alpha _1) / F(B^*) \]

since $\alpha _1$ is algebraic over $F(B^*)$. Now we claim that $B^*$ (counting multiplicity of elements) is algebraically independent over $F$ because if it weren't, then there would be an irreducible $g\in F[x, y_2, \ldots , y_ n]$ such that $g(\beta _1, \alpha _2, \ldots , \alpha _ n) = 0$ which must involve $x$ making $\beta _1$ algebraic over $F(\alpha _2, \ldots , \alpha _ n)$ which would make $\alpha _1$ algebraic over $F(\alpha _2, \ldots , \alpha _ n)$ which is impossible. So this means that $\{ \alpha _2, \ldots , \alpha _ n\} $ and $\{ \beta _2, \ldots , \beta _ m\} $ are bases for $E$ over $F(\beta _1)$ which means by induction, $m = n$. $\square$

Comments (9)

Comment #578 by Wei Xu on

A tpyo in "Now we claim is algebraic over ". I think should be .

Comment #579 by Wei Xu on

A typo in "Now we claim is algebraic over ". I think should be .

Comment #580 by Keenan Kidwell on

In the statement of 030F, should be and should be .

Comment #3178 by Mengxue Yang on

In the second paragraph of the proof, should the extensions etc be instead?

Comment #7386 by Zhenhua Wu on

In the second paragraph of the proof, we should mention that . And I suggest that we should provide a proof for the fact , actually I just prove which will suffice.

Proof: Using axiom of choice, we can construct an injection where denotes the disjoint union, by taking an element into one of the 's that contains it.

There are injections .

Thus and the result follows.

Comment #7387 by Laurent Moret-Bailly on

The fourth sentence of the proof (To prove this...) is unclear. Suggestion: "...consider the collection of algebraically independent subsets such that ".

There are also:

  • 9 comment(s) on Section 9.26: Transcendence

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 030F. Beware of the difference between the letter 'O' and the digit '0'.