Valle Volcanes III
Nota 24 Partes Relacionadas (continuación)
Definition 2.108. Let E/F be a field extension and S ⊆ E. Then S is algebraically dependent over F if there exists s1, ..., sn ∈ S and f (x1, ..., xn) ∈ F [x1, ..., xn] \ {0} such that f (s1, ..., sn) = 0. Otherwise, we say S is algebraically independent over F.
Remarks.
1. ∅ is algebraically independent over any field.
2. {u} is algebraically independent if and only if u is transcendental over F.
3. {s1, ..., sn} is algebraically independent over F if and only if F [s1, ..., sn] ∼= F [x1, ..., xn], where x1, ..., xn are variables.
Lemma 2.109. Let E/F be a field extension and S ⊆ E an algebraically independent set over F. Let u ∈ E. Then S ∪ {u} is algebraically independent if and only if u is transcendental over F (S).
Proof. (⇐) It is enough to show {s1, ..., sn, u} is algebraically independent for s1, ..., sn ∈ S. Suppose f (x1, ..., xn+1) ∈ F [x1, ..., xn+1] and f (s1, ..., sn, u) = 0. Let g(xn+1) = f (s1, ..., sn, xn+1) ∈ F (S)[xn+1]. Note g(u) = 0. Since u is transcendental over F (S), we must have g(xn+1) = 0. Write
Then 0 = g(xn+1) says hi(s1, ..., sn) = 0 for all i. Since {s1, ..., sn} are algebraically independent, we must have hi(x1, ..., xn) = 0 for all i. Thus f (x1, ..., xn+1) = 0.
(⇒) Suppose u is algebraic over F (S). Then u is algebraic over a finite subset of S. So WLOG, S is finite. Then there exists f (x) ∈ F (S)[x] \ {0} such that f (u) = 0. Say
f (x) = gr(s1, ..., sn) hr(s1, ..., sn)
xr+ . . . + g0(s1, ..., sn) h0(s1, ..., sn) ,
where gi(x1, ..., xn), hi(x1, ..., xn) ∈ F [x1, ..., xn]. Multiply f by h0· · · hr to clear denominators and still get a polynomial that u satisfies. So WLOG, hi= 1. Let `(x1, .., xn, x) = gr(x1, ..., xn)xr+ . . . + g0(x1, ..., xn). Note that `(s1, ..., sn, u) = 0. Since S ∪ {u} is algebraically independent, `(x1, ..., xn, x) = 0, a contradiction as f (x) 6= 0. Thus u is transcendental over F (S).
Definition 2.110. Let E/F be a field extension. A set S ⊆ E is called a transcendence base for E/F if S is algebraically independent over F and E/F (S) is algebraic.
Theorem 2.111. Let E/F be a field extension and L ⊆ E an algebraically independent set over F. Then there exists a transcendence base S for E/F such that L ⊆ S.
Proof. Let Γ = {T |L ⊆ T ⊆ E and T is algebraically independent over F }. Note L ∈ Γ so Γ 6= ∅. Let C be any totally ordered subset of Γ. Then T0 = ∪t∈CT ∈ Γ is an upper bound. By Zorn’s Lemma, there exists a maximal set S ∈ Γ. Then S is algebraically independent by definition of Γ and E/F (S) is algebraic by the lemma and maximality of S. Example. Let X, Y be indeterminants over F. Then {X, Y } is a transcendence base for F (X, Y )/F. Also {X2, Y2} is
a transcendence base.
Theorem 2.112. Let E/F be a field extension. Then any two transcendence bases for E/F have the same cardinality. Proof. We’ll prove this in the case that E/F has a finite transcendence base S = {s1, ..., sn}. Let T be a transcendence base for E/F.
Claim: There exists t1∈ T such that {t1, s2, ..., sn} is algebraically independent over E/F.
Proof: Suppose not. Therefore F (T ) is algebraic over F (s2, ..., sn). But E/F (T ) is algebraic, which implies E/F (s2, ..., sn) is, so s1∈ E is algebraic over F (s2, ..., sn), a contradiction.
Claim: The set {t1, s2, ..., sn} is a transcendence base of E/F.
Proof: Suppose s1 is transcendental over F ({t1, s2, ..., sn}). Then {t1, s1, ..., sn} is algebraically independent, but t1 is algebraic over F ({s1, ..., sn}), a contradiction. Thus s1 is algebraic over F ({t1, s2, ..., sn}) which implies F ({t1, s1, ..., sn}) is algebraic over F ({t1, s2, ..., sn}). But E is algebraic over F ({t1, s1, ..., sn}) (as it is over F ({s1, ..., sn})) and thus E is algebraic over F ({t1, s2, ..., sn}).
Repeating this process, replace s2, ..., sn by t2, ..., tn ∈ T to obtain a transcendence base {t1, ..., tn} for E/F. Since T is algebraically independent, T = {t1, ..., tn}.
Definition 2.113. The transcendence degree of E/F is the cardinality of any transcendence base for E/F. Note. The transcendence degree of E/F is 0 if and only if E/F is algebraic.
Theorem 2.114. Suppose K ⊆ F ⊆ E are fields. The tr deg E/K = tr deg E/F + tr deg F/K.
Proof. Let S, T be transcendence bases for E/F and F/K respectively. Since T ⊆ F and S ⊆ E \ F, we see S ∩ T = ∅. Then it is enough to show S ∪ T is a transcendence base for E/K.
Proof: We know that F is algebraic over K(T ). So F (S) is algebraic over K(T )(S) = K(S ∪ T ). As E is algebraic over F (S), E is algebraic over K(S ∪ T ).
Claim 2: S ∪ T is algebraically independent over K.
Proof: Let f (x1, ..., xm, y1, ..., yn) ∈ K[x1, ..., xm, y1, ..., yn] such that f (s1, ..., sm, t1, ..., tn) = 0. We want to show f = 0. Say f = Pgi(y1, ..., yn)hi(x1, ..., xm) where gi ∈ K[y1, ..., yn] and the hi are distinct monomials in the x0s. Let `(x
1, ..., xm) = f (x1, ..., xm, t1, ..., tn) ∈ K(T )[x] ⊆ F [x1, ..., xm]. That that `(s1, ..., sm) = 0. As S is algebraically independence over F, we know ` = 0. So f (x1, ..., xm, t1, ..., tn) = 0. Since the hi(x1, ..., xm) are linearly independent over F [x] (as they are distinct monomials), we must have that gi(t1, ..., tn) = 0 for all i. Since T is algebraically independent over K, gi(y1, ..., yn) = 0. Thus f = 0.
3
Rings and Modules
We will take all rings to have identity, but not necessarily be commutative.
Definition 3.1. Let G be a group, k a field. Let B be a k−vector space with basis {eg}g∈G. Then V is a group ring with elements of the form Pg∈Gcgeg where all but finitely many terms are zero. Define multiplication in V by (Pcgeg)(
P
dgeg) = P
cgdg0egg0.
Remarks. Under this definition, V is a ring with identity element e1. For convenience, we will write g for eg and K[G] for the ring V. Note that K[G] is commutative if and only if G is abelian.
Example. Let G = Cn =< g > and K be any field. Then K[Cn] = { Pn−1
i=0 cigi|ci∈ K}. Define a ring homomorphism K[x] → K[Cn] such that k 7→ k and x 7→ g. Clearly, this is surjective. As gn= 1, we see xn− 1 ∈ ker φ. So we have an induced map K[x]/(xn− 1) → K[C
n]. Since both of these have dimension n, we see that they are isomorphic. Definition 3.2. A division ring is a ring in which every nonzero element is a unit.
Examples.
1. Any field is a division ring.
2. Consider the ring homomorphism R → M2(C) defined by r 7→ rI. In this way, we can consider R as a subring
of M2(C). Let i = " i 0 0 −i # , j = " 0 1 −1 0 # , k = " 0 i i 0 #
. Then {1, i, j, k} are linearly independent over R. Let H = R · 1 + Ri + Rj + Rk ⊆ M2(C). Then H has dimension 4. Note that i2= j2 = k2 = −1, ij = j = −ji, jk =
i = −kj, ki = j = −ik. Thus H is closed under multiplication and has identity. Since H is a vector space, its an additive group. Thus H is a noncommutative subring of M2(C), called the ring of (real) quaternions. Let
α = r0+ r1i + r2j + r3k and α = r0− r1i − r2j − r3k. One can check αα = αα = r20+ r12+ r22+ r23=: |α|2. Note
α = 0 if and only if |α| = 0. So if α 6= 0, α−1= α
|α|2. Thus H is a division ring (but not a field!).
Definition 3.3. Let R be a ring. A left (respectively, right) R-module is an abelian group (M, +) together with a map R × M → M defined by (r, m) 7→ rm such that
1. r(m + n) = rm + rn 2. (r + s)m = rm + sm 3. r(sm) = (rs)m 4. 1m = m
Notes. Not everyone requires (4). In this case, R is called a unital module. Also, we will assume 1 7→ 1 in a ring homomorphism.
Definition 3.4. Let f : R → S be a ring homomorphism such that f (R) ⊆ Z(S). Then S is called an R−algebra. Note. The ker f is a two-sided ideal. Thus f : R/ ker f → S is injective. Thus R/ ker f is commutative and R/ ker f ⊆ Z(S).
Examples. Assume R is a commutative ring.
1. Let R[x1, ..., xn] be the polynomial ring in x1, ..., xn and I an ideal of R[x1, ..., xn]. Then f : R → R[x1, ..., xn]/I defined by r 7→ r = r + I is a ring homomorphism. Thus R[x1, ..., xn]/I is an R−algebra.
2. Define f : R → Mn(R) by r 7→ rI. This is a ring homomorphism, so Mn(R) is an R−algebra.
3. Let G be a group. Define f : R → R[G] by r 7→ re1. This is a ring homomorphism, so R[G] is an R−algebra.
4. Let C(R) = {f : R → R|f is continuous}. Then f : R → C(R) defined by r 7→ fr(x) = r is a ring homomorphism. Thus C(R) is an R−algebra.
Definition 3.5. Let S be a ring, A ⊆ Z(S) a subring, T a subset of S. Say S is generated over A by T if every element of S is a finite sum of elements of the form atn11 · · · tnk
k , where a ∈ A, ti∈ T, ni≥ 0. We write S = A[T ]. If S = A[T ] for some finite subset T of S, then S is finitely generated over A as a ring. If f : R → S is a ring homomorphism with f (R) ⊆ Z(S), then S is a finitely generated R−algebra if S is finitely generated over f (R) as a ring.
Notes.
• If E/K is a finitely generated field extension and F is an intermediate field, then F/K is a finitely generated field extension (HW).
• This is NOT true for algebras. For example, K[x, y] is finitely generated as a K−algebra, but K[x, xy, xy2, ...] is
not finitely generated as a K−algebra. Examples. Let R be a commutative ring.
1. S = R[x1, ..., xn]/I is a finitely generated R−algebra where T = {x1, ..., xn}. Using the above notation, we can say S = R[x1, ..., xn].
Claim. Let S be a finitely generated A−algebra which is commutative. Say S = A[T ] where T = {t1, ..., tn}. Define φ : A[x1, ..., xn] → S by f (x1, ..., xn) 7→ f (t1, ..., tn). Because the t0is commute, φ is an onto ring homomorphism. So S ∼= A[x1, ..., xn]/I.
2. S = Mn(R). Let Eij be the n × n matrix with a 1 in the i, jth entry and zeros everywhere else. Then for A = (aij) ∈ S, we see A =
P
aijEij. Thus S is generated by Eij. So S = R[{Eij}].
3. R[G] is a finitely generated R−algebra if and only if G is a finitely generated group. For one direction, we see if G =< g1, ..., gn >, then R[G] = R[g1, ..., gn].
4. C(R) is not a finitely generated R−algebra.
Let A be a ring. By an A−module, we mean a left A−module, unless when explicitly stated otherwise.
Remark. Let f : R → S be a ring homomorphism. Any S−module M is an R−module via the action r · m := f (r)m. In particular, S is an R−module.
Definition 3.6. Let M be an R−module and T ⊆ M. Say T generates M as an R−module if every element of M can be expressed asPn1riti, for ti∈ T, ri ∈ T, that is, M = RT = R−submodule of M generated by T. We say M is finitely generated as an R−module if M = RT for some finite subset T of M. In practice, if T = {t1, ..., tn}, we will write M = Rt1+ ... + Rtn. Sometimes, this is stated as “M is a finite R−module” even though M is not necessarily finite. Examples. Let R be a commutative ring.
1. R[x1, ..., xn]/I need not be a finitely generated R−module. For example k[x, y]/(xy) is not a finitely generated k−module.
2. Mn(R) is a finitely generated R−module (Mn(R) = P
REij). 3. R[G] is a finitely generated R−module if and only if |G| < ∞.