7.1 Modules

1.
a. 0x = (0 + 0)x = 0x + 0x, so 0x = 0.
c. Using (a), x + (− 1)x = (1 + (− 1))x = 0x = 0 = x + (− x) . Now “subtract” x from both sides by adding −x to both sides.
2.
a. If α : MN is onto and R-linear, and if M = Rx1 + img + Rxn, then N = (x1) + img + (xn) . Since some of the α(xi) may be zero, the result follows.
c. Let K = Rx1 + img + Rxm and let M/K = R(y1 + K) + img + R(yn + K) where the xi and yj are in M. If x img M let

img

with ri img R for each i. Then x − (r1y1 + img + rnyn) is in K, so

img

where sj R for each j. Hence {x1, . . ., xm, y1, . . ., yn} generates M.
3. If Re is an ideal ...

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