Advanced Algebra I
transcendental extension
Before we move on to the transcendental extension. We first com- plete the proof of the Corollary of last time.
Corollary 0.1. Let F/K be an algebraic extension with char(K) = p 6= 0. We have
(1) If F/K is separable, then F = KFpn for each n ≥ 1.
(2) If F/K is finite and F = KFp, then F/K is separable.
(3) In particular, u ∈ F is separable over K if and only if K(up) = K(u).
Note that Fp is not necessarily an extension over K. So is Fpn. But we can take KFpn, which is an extension over K.
Proof. We first suppose that F/K is finite, hence finitely generated.
Write F = K(u1, ..., ur). It’s clear that there is N ≥ 1 such that upN ∈ S. Hence FpN ⊂ S, therefore, KFpN ⊂ S.
We claim that S = KFpN. To see this, one notices that F is purely inseparable over KFpN, so is S purely inseparable over KFpN. And on the other hand, S is separable over K, so is over KFpN. Hence S = KFpN.
For (1), if F/K is separable and finite, then we have F = KFpN. However, in the proof, one can choose N to be arbitrary large. More precisely, one has F = KFpN for all N ≥ N0. By looking at the inclusion
F = KFpN ⊂ KFpN −1 ⊂ ... ⊂ KFp ⊂ F.
One has F = KFpn for all n ≥ 1.
Suppose now that F/K is separable but not necessarily finite. For any u ∈ F , we consider F0 := K(u) which is separable and finite over K. Thus u ∈ F0 = KF0pn ⊂ KFpn for all n ≥ 1. This proves (1).
We now prove (2). If F = KFp, then F = K(KFp)p = KFp2. Inductively, one has F = KFpn for all n ≥ 1. Since we have show that S = KFpN, it follows that F = S.
Apply the statement to a single element. We consider F = K(u).
Fp ⊂ Kp(up) ⊂ K(up) . Indeed, KFp = K(up). By (2), if K(u) = K(up), then u is separable. By (1), if u is separable, then K(u) =
K(up). ¤
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.
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Definition 0.2. 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 0.3. For any u ∈ F , {u} is algebraically dependent over K if and only if u is algebraic over K.
Example 0.4. 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 0.5. 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 0.6. 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 0.7. 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. ¤
Corollary 0.8. 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
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directly that S ∪ {u} is algebraically dependent if u ∈ K(S). Thus S is a maximal algebraically independent set. ¤ Corollary 0.9. 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 0.10. 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 0.11. Let S, T be transcendental bases of F/K. If S is infinite, then |T | = |S|.
Proof. By the previous theorem, we have |T | is infinite.
For each s ∈ S, s is algebraic over K(T ). There is a finite subset Ts 6= ∅ ⊂ T containing all coefficients of the minimal polynomial of s.
And hence s is algebraic over K(Ts). Let T0 := ∪s∈STs. Since every u ∈ F is algebraic over K(S) and hence algebraic over K(T0). It follows that T0 = T as T0 ⊂ T .
Finally, one shows that
|T | = | ∪s∈S | ≤ |S||N| = |S|.
Since one can similarly have |S| ≤ |T |. We are done. ¤ With the above two theorem, we can have a well-defined notion of transcendental degree.
Definition 0.12. Let F/K be an extension. The transcendental degree of F/K, denoted tr.d.F/K, is the cardinal number |S|, where S is a transcendental basis.
Theorem 0.13. If F/E and E/K are extensions, then tr.d.F/K = tr.d.F/E + tr.d.E/K.
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Proof. Let S be a transcendental basis of E/K and T be a transcen- dental basis of F/E. It clear that T ∩ E = ∅, thus T ∩ S = ∅. It’s enough to show that S ∪ T is a transcendental basis of F/K.
Note that E is algebraic over K(S), so E is algebraic over K(S ∪ T ).
It follows that E(T ) is algebraic over K(S ∪ T ). Together with the fact that F is algebraic over E(T ). One sees that F is algebraic over K(S ∪ T ).
It suffices to show that S ∪ T is algebraically independent. If f is a non-trivial algebraic relation, i.e. f (s1, ..., sn, t1, ..., tm) = 0 for some
si ∈ S, tj ∈ T . ¤
Corollary 0.14. Let F1/K1 and F2/K2 are extensions and F1, F2 are algebraically closed. Then every isomorphism between K1 and K2 can be extended to an isomorphism between F1 and F2 if tr.d.F1/K1 = tr.d.F2/K2
