By Randall R. Holmes

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**Extra resources for Abstract Algebra II**

**Example text**

Let S ∈ A . We need to show that ϕ(ϕ−1 (S )) = S . Let s ∈ S . Since ϕ is surjective, there exists r ∈ R such that ϕ(r) = s . But this last equation says that r ∈ ϕ−1 (S ), so s ∈ ϕ(ϕ−1 (S )). This gives S ⊆ ϕ(ϕ−1 (S )). The other inclusion is immediate. (b) Let S, T ∈ A. If ϕ(S) ⊆ ϕ(T ), then, using (a), we have S = ϕ−1 (ϕ(S)) ⊆ ϕ−1 (ϕ(T )) = T, and the other implication is immediate. Assume that S ⊆ T . We claim that the map f : {t + S | t ∈ T } → {ϕ(t) + ϕ(S) | t ∈ T } given by f (t + S) = ϕ(t) + ϕ(S) is a well-defined bijection.

In more detail, V /W = {v + W | v ∈ V } is a vector space with addition and scalar multiplication given by (i) (v + W ) + (v + W ) = (v + v ) + W , (ii) a(v + W ) = (av) + W . These operations are well-defined, meaning that they do not depend on the choices of coset representatives. V /W is the quotient of V by W . 6 Span Let F be a field, let V be a vector space over F , and let S be a subset of V . The span of S, written S , is the intersection of all subspaces of V that contain S: W. S = W ≤V W ⊇S Since an intersection of subspaces is again a subspace, the span of S is a subspace of V .

We say that r divides s, written r|s, if s = ra for some a ∈ R. We say that r and s are associates, written r ∼ s, if r = su for some unit u ∈ R. As the notation suggests, the property of being associates is an equivalence relation on R. Here are some examples: • Let r ∈ R. Then r|0. However, 0|r if and only if r = 0. • Let R = Z. Then 2|6 since 6 = (2)(3). For any integer n, we have n ∼ −n, since n = n(−1) and −1 is a unit. In fact, for n, m ∈ Z, we have n ∼ m if and only if m = ±n. • Let R = Q.