Advanced Algebra II
transcendental extension
We now start our discussion on transcendental extension. The main purpose is to show that the concept of transcendental degree, which is the cardinality of transcendental basis, can be well-defined. Moreover, transcendental degree is a good candidate for defining dimension.
Definition 0.1. Let F/K be an extension. S ⊂ F is said to be al- gebraically dependent (over K) if there is an n ≥ 1 and an f 6= 0 ∈ K[x1, ..., xn] such that f (s1, ..., sn) = 0 for some s1, ..., sn. Roughly speaking, some element of S satisfy a non-zero algebraic relation f over K.
S is said to be algebraically independent over K if it’s not alge- braically dependent over K.
Example 0.2. For any u ∈ F , {u} is algebraically dependent over K if and only if u is algebraic over K.
Example 0.3. In the extension K(x1, ..., xn)/K, S = {x1, ..., xn} is algebraically independent over K.
The following theorem says that finitely generated purely transcen- dental extension are just rational function fields.
Theorem 0.4. If {s1, ..., sn} ⊂ F is algebraically independent over K.
Then K(s1, ..., sn) ∼= K(x1, ..., xn).
Proof. We consider the homomorphism θ : K[x1, ..., xn] → K[s1, ..., sn].
θ is surjective by definition. It’s injective because {s1, ..., sn} ⊂ F is algebraically independent. Then θ induces an isomorphism on quotient
fields. ¤
One notices that the notion of being algebraic independent is an analogue of being linearly independent. Therefore, one can try to define the notion of ”basis” and ”dimension” in a similar way.
Definition 0.5. S ⊂ F is said to be a transcendental basis of F/K if S is a maximal algebraically independent set. In other words, for all u ∈ F − S, S ∪ {u} is algebraically dependent.
We will then define the transcendental degree to be the cardinality of a transcendental basis (in a analogue of dimension). In order to show that this is well-defined. We need to work harder.
Proposition 0.6. Let S ⊂ F be an algebraically independent set over K and u ∈ F − K(S). Then S ∪ {u} is algebraically independent if and only if u is transcendental over K(S).
Proof. The proof is straightforward. ¤
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Corollary 0.7. S is a transcendental basis of F/K if and only if F/K(S) is algebraic.
Proof. Suppose that S is a transcendental basis of F/K. If u ∈ F − K(S), then S ∪ {u} is not algebraically independent. Thus, u is alge- braic over K(S) by the Proposition.
On the other hand, suppose that F/K(S) is algebraic. Then for all u ∈ F − S, u is algebraic over K(S). By the Proposition, S ∪ {u}
is algebraically dependent if u ∈ F − K(S). In fact, it’s easy to see directly that S ∪ {u} is algebraically dependent if u ∈ K(S). Thus S is a maximal algebraically independent set. ¤ Corollary 0.8. Let S ⊂ F be an subset over such that F/K(S) is algebraic. Then S contains a transcendental basis.
Proof. By Zorn’s Lemma, there exists a maximal algebraically inde- pendent subset S0 ⊂ S. Then K(S) is algebraic over K(S0) and hence
F is algebraic over K(S0). ¤
Theorem 0.9. Let S, T be transcendental bases of F/K. If S is finite, then |T | = |S|.
Proof. Let S = {s1, ..., sn} and S0 := {s2, ..., sn}. We first claim that there is an element t ∈ T , say t = t1 such that {t1, s2, ..., sn} is a transcendental basis.
to see this, if every element of T is algebraic over K(S0), then F is algebraic over K(T ) hence over K(S0) which is a contradiction. Thus, there is an element t ∈ T , say t = t1such that t1 is transcendental over K(S0). And hence T0 := {t1, s2, ..., sn} is algebraically independent.
By the maximality of S, one sees that s1 is algebraic over K(T0).
It follows that F is algebraic over K(t1, s1, ..., sn) and hence algebraic over K(T0). Therefore, T0 is a transcendental basis.
By induction, one sees that there is a transcendental basis {t1, ..., tn} ⊂
T . Thus T = {t1, ..., tn}. ¤
Theorem 0.10. Let S, T be transcendental bases of F/K. If S is infinite, then |T | = |S|.
Proof. By the previous theorem, we may assume that T is infinite as well.
For s ∈ S, we have s ∈ F hence algebraic over K(T ). Let Ts⊂ T be the subset of T of elements that appearing in the minimal polynomial of s. It’s clear that Ts 6= ∅ otherwise, s is algebraic K which is not the case. Also note that Ts is finite.
Let T0 := ∪s∈STs. We claim that T0 = T . To this end, one sees that for u ∈ F , u is algebraic over K(S) and hence algebraic over K(T0).
Thus F/K(T0) is algebraic. T is a transcendental basis, hence T = T0. Lastly, one sees that
|T | = |T0| = | ∪s∈S Ts| ≤ |S| · ℵ0 = |S|.
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Replace S by T , one has |S| ≤ |T |. We are done. ¤ With these two theorem, we can define the transcendental degree of an extension. And the definition is independent of choices of basis.
Definition 0.11. Let F/K be an extension and S be a transcendental basis. We define the transcendental degree of F/K, denoted tr.d.F/K, to be |S|.
Theorem 0.12. Let F/E and E/K be extensions. Then tr.d.F/K = tr.d.F/E + tr.d.E/K.
Proof. Let T be a transcendental basis of F/E and S be a transcen- dental basis of E/K. We would like to show that S ∪ T is a transcen- dental basis of F/K. Note that T ∩ E = ∅, hence S ∩ T = ∅. Thus
|S ∪ T | = |S| + |T |, and we are done.
To see the claim, it’s easy to check that E(T ) = EK(S ∪ T ). Hence E(T )/K(S ∪ T ) is algebraic if E/K(S) is algebraic. Also, F/E(T ) is algebraic, therefore, F/K(S ∪ T ) is algebraic.
It suffices to show that S ∪ T is algebraically independent. Suppose that there is f (x1, ..., xn, y1, ..., ym) such that f (s1, ..., sn, t1, ..., tm) = 0.
We can write
f (x1, ..., xn, y1, ..., ym) =X
I
hI(x1, ..., xn)yI,
and we have P
IhI(s1, ..., sn)tI. Since T is algebraically independent over E 3 hI(s1, ..., sn). It follows that hI(s1, ..., sn) = 0 for all I. Since S is algebraically independent over K, if follows that hI(x1, ..., xn) = 0 ∈ K[x1, ..., xn] for all I. Therefore f (x1, ..., xn, y1, ..., ym) = 0. Hence
S ∪ T is algebraically independent. ¤
