6.2% (F51) Sep, .19, 2016
1\.$olution. We claim that. the product and coproduct. of the family of a
single object (01-) are both 01 itself (up to isomorphism).
First the projection p; for the product 01 is the identity map 19. Actuall
ber einem Ring
Eine Abbildung f : R S zwischen zwei Ringen R und S heit Ringhomomorphismus, wenn f
ur a, b R gilt:
f (a + b) = f (a) + f (b)
f (a b) = f (a) f (b)
f (1) = 1.
Beispiele. 1. Eine Teilmenge R ein
59p. 24. 2016
1.301ution. (a) If (1.5 are isomorphic. he > (1.!) >.- rt. it follows that at S b.
b' 6 o, by the denition of morphisms in the ategory 9. Moreover, n is a poset.
So a; = b-
(b). By the denition, 1.I : a -3- a is one and
1. Let G be a group. If a G and a a = a , show that a = 1.
Proof. As G is a group, every element in G has an inverse element a1 .
a a = a implies a = a 1 = a (a a1 ) = (a a) a1 = a a1 = 1.
2. Let G, H be groups. If f : G H is a bijective homomorphis
1. Let R, S be rings with |R| = |S| = p, p prime. Then R
= S as rings.
Proof. There is (exactly) one ring homomorphism c : Z R by Proposition
1. It follows that (p) Ker(c) = (n) by Lagrange theorem, which yields 1 = n|p
since p is prime, and thus n