470–2 Winter 2017
Algebra II
Homework 3
Due: Feb 26, 2017
All rings are commutative and nonzero.
1. [300 ] Consider a commutative diagram of R-modules and homomorphisms such that
each row is exact:
/M
/ M 00
/ 0.
M0
0
/
f
/
N0
g
N
h
/ N 00
Prove:
(a) If f, h are injective then g is injective.
(b) If f, h are surjective then g is surjective.
(c) Assume in addition that 0 → M 0 → M is exact and that N → N 00 → 0 is exact.
Prove that if any two of f, g, h are isomorphisms, then so it the third.
[Hint: Use the stake lemma.]
2. [100 ] Let M, N be two flat R-modules. Show that M ⊗R N is flat.
3. [100 ] Let k be a field, f (X) an irreducible polynomial over k, and α a root of f . Show
that for every field extension k 0 of k, we have that k(α) ⊗k k 0 is isomorphic, as a
k 0 -algebra, to k 0 [X]/(f (X)).
4. [100 ] Let E be a finite extension of a field k. Show that E is separable over k if and
only if E ⊗k L has no (nontrivial) nilpotent elements for all field extensions L over k.
5. [400 ] Let M be an R-module. We say that M is faithfully flat if M is flat, and such
that E 6= 0 implies E ⊗R M 6= 0 for every R-module E. Prove that the following
conditions are equivalent.
(a) M is faithfully flat.
(b) M is flat, and for all maximal ideals m of R, we have mM 6= M .
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(c) M is flat, and if u : F → E is a nonzero homomorphism of R-modules, then the
induced homomorphism ũ : F ⊗R M → E ⊗R M is also nonzero.
(d) A sequence of R-modules N 0 → N → N 00 is exactly if and only if N 0 ⊗R M →
N ⊗R M → N 00 ⊗R M is exact.
[Hint: Show that (a)⇒(b)⇒(c)⇒(d)⇒(a). Of course, you may try other directions.]
Solutions.
1. We have the diagram
M0
0
/
u0
/
M
f
N0
v0
/
/
u
g
N
v
/
M 00
/
0.
h
N 00 .
For instructional reason, I will give direct proofs for (a) and (b), which are part of the
snake lemma we skipped in class. For your solution, it is fine if you use the snake
lemma directly.
(a) Take m ∈ ker(g). Put m00 = u(m). Then h(m00 ) = h(u(m)) = v(g(m)) = 0.
Since h is injective, m00 = 0. Thus there is m0 ∈ M 0 such that m = u0 (m0 ). Then
v 0 (f (m0 )) = g(u0 (m0 )) = g(m) = 0. Since both f and v 0 are injective, we have m0 = 0.
Thus m = 0 and g is injective.
(b) Take n ∈ N . As h is surjective, there is m00 ∈ M 00 such that v(n) = h(m00 ).
As u is surjective, there is m ∈ M such that m00 = u(m). Now v(n − g(m)) =
v(n) − v(g(m)) = h(m00 ) − h(u(m)) = h(m00 ) − h(m00 ) = 0. Thus there is n0 ∈ N 0 such
that v 0 (n0 ) = n − g(m). Since f is surjective, there is m0 ∈ M 0 such that f (m0 ) = n0 .
Now g(u0 (m0 ) + m) = g(u0 (m0 )) + g(m) = v 0 (f (m0 )) + g(m) = v 0 (n0 ) + g(m) = n −
g(m) + g(m) = n. Thus h is surjective.
(c) The assumption, together with the snake lemma, implies the following exact sequence
0 → ker(f ) → ker(g) → ker(h) → coker(f ) → coker(g) → coker(h) → 0.
Then the rest is trivial.
2. Let F 0 ,→ F be a given inclusion of R-modules. Since M is flat, F 0 ⊗R M → F ⊗R M
is injective. Since N is flat, (F 0 ⊗R M ) ⊗R N → (F ⊗R M ) ⊗R N is injective. But the
last homomorphism is simply F 0 ⊗R (M ⊗R N ) → F ⊗R (M ⊗R N ). Thus M ⊗R N is
flat.
3. Since both k[X] and k 0 are free k-modules, it is easy to see that k[X] ⊗k k 0 ' k 0 [X].
As k 0 is flat over k, k 0 [X] is flat over k[X]. Therefore k[X]/(f (X)) ⊗k[X] k 0 [X] '
k 0 [X]/(f (X)). On the other hand, we have k[X]/(f (X)) ⊗k[X] k 0 [X] ' k[X]/(f (X)) ⊗k
k 0 = k(α) ⊗k k 0 .
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4. Suppose that E/k is separable. Then E = k(α) for some α ∈ E, whose minimal
polynomial F (X) ∈ k[X] is separable. By Problem 3, we have E ⊗k L ' L[X]/(f (X)).
Q
Suppose that f (X) = i fi (X)ei (ei ≥ 1) is the prime decomposition of f (X) in
Q
L[X]. Then ei = 1 as f is separable. Therefore, L[X]/(f (X)) = L[X]/( i fi (X)) '
Q
i=1 L[X]/(fi (X)) by Chinese Remainder Theorem. Since fi (X) is irreducible in L[X],
L[X]/(fi (X)) is a field. Thus, E ⊗k L is a finite product of fields, hence contains no
nilpotent elements.
Suppose that E/k is inseparable. Let p be the characteristic of k, which is a prime.
Then there is an element α ∈ E such that α 6∈ k but αp ∈ k. Take L = E. Consider
β := α⊗1−1⊗α ∈ E ⊗k E. Then β 6= 0 (otherwise, α ∈ k). But β p = (α⊗1−1⊗α)p =
αp ⊗ 1 − 1 ⊗ αp = 0 as αp ∈ k. Thus E ⊗k E contains (nontrivial) nilpotent elements
(such as β above).
5. (a)⇒(b): Since M is flat, M/mM ' R/m ⊗R M . If M is faithfully flat, then R/m ⊗R
M 6= 0 as R/m 6= 0. Thus mM 6= M .
(b)⇒(c): First M is flat, as it is said in (b). Let u : F → E be a nonzero homomorphism. Take y ∈ F such that u(y) 6= 0. Define a homomorphism v : R → E such that
v(a) = u(ay) for a ∈ R. Then a : ker(v) is a proper ideal of R. We have an exact
v
sequence 0 → a → R →
− E. Since M is flat, the induced sequence
ṽ
0 → a ⊗R M → R ⊗R M →
− E ⊗R M
is exact. In particular, we have a monomorphism ṽ : M/aM ,→ E ⊗R M . Let m be
a maximal ideal containing a. Since M/mM 6= 0, M/aM 6= 0 hence im(ṽ) 6= 0. But
since ṽ factors through ũ, we have im(ũ) 6= 0, i.e., ũ is also nonzero.
(c)⇒(d): “Only if” direction: (For this part, we only need that M is flat.) Consider
the sequence
f
g
N0 →
− N→
− N 00 .
(1)
If it is exact, then it induces another short exact sequence 0 → N 0 / ker(f ) → N → N 00 .
Since M is flat, the induced sequence
0 → (N 0 / ker(f )) ⊗R M → N ⊗R M → N 00 ⊗R M
is exact. Since N 0 ⊗R M → (N 0 / ker(f )) ⊗R M is surjective, N 0 ⊗R M → N ⊗R M →
N 00 ⊗R M is exact. “If” direction: Suppose that the induced sequence
f˜
g̃
N 0 ⊗R M →
− N ⊗R M →
− N 00 ⊗R M
is exact. In particular, the composite homomorphism N 0 ⊗R M → N 00 ⊗R M is zero.
By (c), the homomorphism g ◦ f : N 0 → N 00 is zero as well, i.e., (1) is a complex. It
remains to show that ker(g)/ im(f ) = 0. We always have (ker(g)/ im(f )) ⊗R M '
(ker(g) ⊗R M )/(im(f ) ⊗R M ) and im(f ) ⊗R M ' im(f˜). Since M is flat, we also have
ker(g)⊗R M ' ker(g̃). Therefore, the identity homomorphism (ker(g)/ im(f ))⊗R M →
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(ker(g)/ im(f )) ⊗R M is isomorphic to the identity homomorphism ker(g̃)/ im(f˜) →
ker(g̃)/ im(f˜). But the latter is zero homomorphism as ker(g̃) = im(f˜). Thus by (c),
ker(g)/ im(f ) is zero.
(d)⇒(a): Let E be an arbitrary R-module. If E ⊗R M is zero, then 0 ⊗R M →
E ⊗R M → 0 ⊗R M is exact. By (d), 0 → E → 0 is exact, i.e., E = 0.
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