The objective of this article is to establish the theory of transcendence bases for field extensions. Readers who are already familiar with this may skip the article. We will focus on field extensions here.
Let be a field extension. Elements are said to be algebraically dependent over K if
Otherwise, are algebraically independent over K.
We say forms a transcendence basis of L over K if it is algebraically independent over K and L is algebraic over .
If such a basis exists for , we say L has finite transcendence degree over K, and write
the transcendence degree of L over K.
Prove that if then so forms a transcendentce basis of L over .
Note that a priori, it is not clear that the transcendence basis is well-defined: could we perhaps choose two transcendence bases of different sizes? The answer is no as we will see.
For write for the field extension of K generated by .
If are algebraically independent over K, then every element of L’ can be uniquely written as as runs through all rational functions . Thus we have an isomorphism:
Consider the K-algebra homomorphism which takes for each . By definition of algebraic independence this map is injective so we get an embedding
of fields, whose image is clearly L’. ♦
The following will be repeatedly used throughout the article.
Let be a field extension and be algebraically independent over K. Let .
Then are algebraically dependent over K if and only if is algebraic over .
Compare this with the linear algebra variant: if V is a vector space and are linearly independent, then for , are linearly dependent if and only if w is a linear combination of .
(⇒) If are algebraically dependent over K we have a non-zero polynomial such that . This polynomial must involve Y or we would obtain a polynomial relation in , a contradiction. Taking
we see that and . ♦
(⇐) If is algebraic over pick , not all zero, such that
Clearing denominators, we assume each so for some non-zero polynomial with coefficients in K. ♦
Here is the main result.
Let be a transcendence basis of L over K and be transcendental (i.e. not algebraic) over K. Then there exists an such that
form a transcendence basis of L over K.
Furthermore if are algebraically independent over K, then we can pick from .
By assumption is algebraic over . Thus by the key observation, are algebraically dependent. Assume
- are algebraically independent but
- are algebraically dependent.
We claim that swapping with also gives a transcendence basis.
Indeed let . By the key observation, is algebraic over and hence over K’. So we have algebraic extensions .
It remains to show that are algebraically independent. If not, by the key observation is algebraic over and we have algebraic extensions , so is algebraic over , a contradiction. ♦
Transcendence Degree is Well-Defined
If is a transcendence basis for and are algebraically independent over K, then
- we can replace m terms of by to obtain another transcendence basis for L over K.
We repeatedly apply swapping lemma. If m = 0, there is nothing to prove; otherwise by swapping lemma, we can swap out some element of with to obtain another transcendence basis.
At each stage, suppose we have a transcendence basis together with n–j elements of . If j = m we are done. If j = n then is already a transcendence basis so we also have j = m. Hence suppose and .
Since are algebraically independent, swapping lemma asserts we can swap out another with so we get a transcendence basis together with n–j-1 elements of . Eventually, we obtain a transcendence basis comprising of with n – m elements of . ♦
Any two transcendence bases of L/K have cardinality n so is well-defined.
For any transcendence bases and , proposition 1 says and . ♦
Let be a field extension and such that L is algebraic over .
Then contains a transcendence basis of L over K.
Pick a maximal algebraically independent subset of . Upon reordering suppose this is ; set . By maximality of , each of is algebraic over . Hence we get algebraic extensions
and we are done. ♦
Let be field extensions. Then M has finite transcendence degree over K if and only if M has finite transcendence degree over L and L has finite transcendence degree over K, in which case
(⇒) Let be a transcendence basis of M over K. In particular M is algebraic over so by lemma 2, there exists a subset of which forms a transcendence basis of M over L.
Next suppose are algebraically independent over K. By proposition 1, so there exists a maximal algebraically independent sequence for L/K. By key observation it follows that any is algebraic over .
(⇐) Pick transcendence bases for L over K, and for M over L. We claim that form a transcendence basis of M over K, which would complete the proof.
Suppose for a non-zero . Then is a polynomial relation for ; since these are algebraically independent we have . But the coefficients of q are polynomials in ; since are algebraically independent, we also get .
Finally, since is algebraic, so is
since the RHS is generated as a field by L and , all algebraic over the LHS. Furthermore, is algebraic by assumption. Thus
is an algebraic extension as desired. ♦
Let where is irreducible. Prove that has transcendence degree n – 1 over K.
Can you please check the proof of Proposition . Especially the statement. “… Hence suppose j>n and j>m….” Also in the preceding statement how does it follow that ?
You’re right it should be j < n and j < m.
Next on your question why j = m. If j < m, then is a transcendence basis so L is algebraic over so is algebraic over which is a contradiction.