Exercise B10

Let C be a Ab-category and f ∈ Hom_{C}(B, C).

1. (f is monic) ⇔ (fe = 0 ⇒ e = 0 ∀e ∈ Hom_{C}(A, B), A ∈ Obj(C)). Proof. ⇒. If f is monic, set fe = 0 = fe−fe = f(e−e) = f0, which implies e = 0. ⇐. Let e_{1}, e_{2} ∈ Hom_{C}(A, B). Set fe_{1} = fe_{2}, wich implies f(e_{1} − e_{2}) = 0, i.e. e_{1} = e_{2}. □

2. (f is epi) ⇔ (gf = 0 ⇒ g = 0 ∀g ∈ Hom_{C}(C, D), D ∈ Obj(C)). Proof. ⇒. if f is epi, set gf = 0 = (g − g)f = 0f which implies g = 0. ⇐. Let g_{1}, g_{2} ∈ Hom_{C}(C, D). Set g_{1}f = g_{2}f wich implies (g_{1} − g_{2})f = 0, i.e. g_{1} = g_{2}. □

C1 Exercise 8A

(i) E_{1} ⊗ E_{2} ≅ E_{2} ⊗ E_{1}. Proof. We define the map E_{1} × E_{2} ⟶^{f} E_{2} ⊗ E_{1} with (x, y) ⟼^{f} y ⊗ x. f is clearly bilinear, therefore it induces a homomorphism E_{1} ⊗ E_{2} ⟶^{f_{∗}} E_{2} ⊗ E_{1} such that f_{∗}(x ⊗ y) = y ⊗ x (using the universal property of the tensor product). Similarly, we have a well defined homomorphism E_{2} ⊗ E_{1} ⟶^{g_{∗}} E_{1} ⊗ E_{2} such that g_{∗}(y⊗x) = x⊗y. Then f_{∗}g_{∗}(y⊗x) = y⊗x and g_{∗}f_{∗}(x⊗y) = x⊗y wich implies f_{∗}g_{∗} = 1_{E_{2}⊗E_{1}} and g_{∗}f_{∗} = 1_{E_{1}⊗E_{2}} since we have equality on the generators. □

(ii) (E_{1} ⊗ E_{2}) ⊗ E_{3} ≅ E_{1} ⊗ (E_{2} ⊗ E_{3}). Proof. Let z ∈ E_{3} fixed. Define the homomorphism E_{2} ⟶^{f_{z}} E_{2} ⊗ E_{3} with f_{z}(y) = y ⊗ z wich induces (by functoriality of E_{1} ⊗ −) a well defined homomorphism

\begin{aligned}g_{z} = 1_{E_{1}} ⊗ f_{z} : & E_{1} ⊗ E_{2} & ⟶ & E_{1} ⊗ (E_{2} ⊗ E_{3}) \\ & x ⊗ y & ⟼ & x ⊗ (y ⊗ z) \end{aligned}

We now define the map

\begin{matrix}(E_{1} ⊗ E_{2}) × E_{3} & ⟶^{φ} & E_{1} ⊗ (E_{2} ⊗ E_{3}) \\ (m, z) & ⟼ & g_{z}(m) \end{matrix}

with m = ∑_{k} x_{k} ⊗ y_{k} a general element of E_{1} ⊗ E_{2}. φ is clearly bilinear and therefore induces the homomorphism

\begin{matrix}(E_{1} ⊗ E_{2}) ⊗ E_{3} & ⟶^{φ_{∗}} & E_{1} ⊗ (E_{2} ⊗ E_{3}) \\ (x ⊗ y) ⊗ z & ⟼ & x ⊗ (y ⊗ z) \end{matrix}

Similarly we construct a well defined homomorphism

\begin{matrix}E_{1} ⊗ (E_{2} ⊗ E_{3}) & ⟶^{ψ_{∗}} & (E_{1} ⊗ E_{2}) ⊗ E_{3} \\ x ⊗ (y ⊗ z) & ⟼ & (x ⊗ y) ⊗ z \end{matrix}

such that φ_{∗}ψ_{∗} = 1_{E_{1}⊗(E_{2}⊗E_{3})} and ψ_{∗}φ_{∗} = 1_{(E_{1}⊗E_{2})⊗E_{3}}. The proof of the isomorphism (E_{1} ⊗ E_{2}) ⊗ E_{3} ≅ E_{1} ⊗ E_{2} ⊗ E_{3} is completely similar. □

C2 Exercise 9A

(1) (ℤ/nℤ) ⊗_{ℤ} (ℤ/mℤ) ≅ ℤ/(n, m)ℤ. Proof. Applying the formula M ⊗_{R} R/I ≅ M/IM with M = ℤ/nℤ , R = ℤ and I = mℤ gives (ℤ/nℤ) ⊗_{ℤ} (ℤ/mℤ) ≅ (ℤ/nℤ)/m(ℤ/nℤ) = G. Since G is the quotient of a finite cyclic group by one of its subgroup, the result is a finite cyclic group (if x generates ℤ/nℤ then \overline{x} generates G). Therefore G = ℤ/dℤ for some d. Note that m(ℤ/nℤ) = {0, m, 2m, ⋯, (n − 1)m} is generated by m in ℤ/nℤ. We denote by k the order of m in ℤ/nℤ, then n = kd. k is the smallest integer such that km = 0 mod n, i.e. the smallest integer s.t. n|km. This is equivalent to be the smallest integer s.t. \frac{n}{(n,m)}|k \frac{m}{(n,m)}. Since (\frac{n}{(n,m)}, \frac{m}{(n,m)}) = 1, this is equivalent to be the smallest integer s.t. \frac{n}{(n,m)}|k, therefore k = \frac{n}{(n,m)}. Since n = kd, we have d = (n, m). □

(2) Let R = k[x, y] and I = (x, y). The element x ⊗ y − y ⊗ x = 0 in R ⊗_{R} R because x ⊗ y = (xy) ⊗ 1 = y ⊗ x. In I ⊗_{R} I, the above operation is not allowed since 1 ∉ I. Furthermore x ⊗ y − y ⊗ x ≠ 0 in I ⊗_{R} I. Proof. Suppose that x ⊗ y − y ⊗ x = 0 in I ⊗_{R} I. Any element i ∈ I = (x, y) can be uniquely expressed in the form i = ax + by + f(x, y) for some a, b ∈ k and f(x, y) ∈ R of degree greater than 2. Therefore we can define a biliniar map

\begin{matrix}I × I & ⟶^{f} & k^{2} ⊗_{R} k^{2} \\ (i, j) & ⟼ & (a, b) ⊗ (c, d) \end{matrix}

for i = ax + by + f(x, y) and j = cx + dy + g(x, y). Therefore f induces, via the UMP of ⊗, a unique R-module homomorphism I ⊗_{R} I ⟶^{f_{∗}} k^{2} ⊗_{R} k^{2} such that

\begin{matrix}f_{∗}(i ⊗ j) = & (a, b) ⊗ (c, d) = (a ⊗ (c, d), b ⊗ (c, d)) \\ = & (a ⊗ c, a ⊗ d, b ⊗ c, b ⊗ d) = (ac, ad, bc, bd) \end{matrix}

using the identifications k^{2} ⊗_{R} k^{2} = k^{2} ⊗_{k} k^{2} ≅ k^{4}. We have f_{∗}(x ⊗ y) = (0, 1, 0, 0) and f_{∗}(y ⊗ x) = (0, 0, 1, 0). By assumption x ⊗ y − y ⊗ x = 0, so f_{∗}(x ⊗ y − y ⊗ x) = (0, 1, 0, 0) − (0, 0, 1, 0) = 0 which is a contradiction, therefore x ⊗ y − y ⊗ x ≠ 0 in I ⊗_{R} I. □