Homework 5 M 373K Solutions
نویسنده
چکیده
m = 0: When m = 0, Z/0 is Z, because every number is equivalent to itself (mod 0), but not to any other. But Z does not have multiplicative inverses, and so it cannot be a field. m = 1: When m = 1, Z/Z is {0}, because every number is equivalent to every other number (mod 1). But {0} does not have a non-zero multiplicative inverse, because 1 ≡ 0, and so {0} is not a field. m ≥ 4: Here, we have that m is composite, and therefore there are at least 2 factors a, b of m such that ab = m, but 1 < a, b < m. Since a, b < m, we have that a 6≡ 0 (mod m) and m 6≡ 0 (mod m), but ab ≡ m ≡ 0 (mod m). Now consider just the elements a, b. We know that in a field, they must both have an inverse. In particular, there is some a−1 ∈ Z/mZ such that a−1a ≡ 1 ≡ aa−1. But then a−1ab ≡ a−10 ⇒ b ≡ 0, which contradicts the fact that 1 < b < m (here note that, in any ring, a · 0 = 0 since a · 0 = a · (0 + 0) = a · 0 + a · 0, then cancel the a · 0). Therefore, a−1 cannot exist, and Z/mZ is not a field.
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تاریخ انتشار 2015