Algebra: Monomials and Polynomials by John Perry

By John Perry

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Extra resources for Algebra: Monomials and Polynomials

Example text

3. 58. We claim that (M, ×) is isomorphic to (N, +). To see why, let f : M −→ N by f ( x a ) = a. First we show that f is a bijection. To see that it is one-to-one, let t , u ∈ M, and assume that f ( t ) = f ( u ). By definition of M, t = x a and u = x b for a, b ∈ N. By definition of f , f ( x a ) = f x b ; by substitution, a = b . In this case, x a = x b , so t = u. We assumed that f ( t ) = f ( u ) for arbitrary t , u ∈ M, and showed that t = u; that proves f is one-to-one. To see that it is onto, let a ∈ N.

This is false! After all, 2 is not an integer power of 3, and 3 is not an integer power of 2. ∼ (N, ×). The claim was correct: (N, +) = ∼ is an equivalence relation; thus, we can also conclude You will show in the exercises that = ∼ that (N, ×) = (N, +). Let’s look again at monomials. It might have occurred to you that we can view any element of Mn as a list of n elements of M. ) If not, here’s an example: x16 x23 looks an awful lot like x 6 , x 3 . We can do this with other sets, as well; creating new sets via lists of elements of old sets is very useful.

N − 1 because the real and imaginary parts do not agree (think of a circle, and see Figure ). Since there can be only n distinct roots, Ω n is a complete list of nth roots of unity. We only sketch the proof that Ω n is a cyclic group. 71 that xy ∈ Ω n . The complex numbers are associative under multiplication; since Ω n C, the elements of Ω n are also associative under multiplication. The multiplicative identity 1 ∈ Ω n since 1n = 1 for all n ∈ N+ . 72 that x −1 ∈ Ω n . 67 tells us that α ∈ Ω n ; the remaining elements are powers of α.

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