By Randall R. Holmes

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B) Give an example to show that if ϕ is not surjective, then ϕ(1) need not be an identity element for R . 6–2 Let R and R be rings and let ϕ : R → R be a homomorphism. (a) Prove that if J is an ideal of R , then ϕ−1 (J) is an ideal of R. (b) Prove that if I is an ideal of R and ϕ is surjective, then ϕ(I) is an ideal of R . (c) Give an example to show that, without the assumption of surjectivity in (b), ϕ(I) need not be an ideal of R . 8). For r ∈ R, define a function λr : R → R by λr (s) = rs.

It is convenient to use this isomorphism to view F as a subring of R. 3 Basic identities Let F be a field and let V be a vector space over F . 1 Theorem. (i) 0v = 0 for all v ∈ V , (ii) a0 = 0 for all a ∈ F , (iii) (−a)v = −av and a(−v) = −av for all a ∈ F, v ∈ V . Proof. (i) Let v ∈ V . Since 0v + 0v = (0 + 0)v = 0v, cancellation gives 0v = 0. (ii) Let a ∈ F . Since a0 + a0 = a(0 + 0) = a0, cancellation gives a0 = 0. (iii) Let a ∈ F and v ∈ V . Since av + (−a)v = (a + (−a))v = 0v = 0 (using part (i)), we have (−a)v = −av.

This completes the proof of the existence statement. 4. Let r ∈ R. Suppose that r has two factorizations, r = s1 s2 · · · sm and r = t1 t2 · · · tn with each si and each ti irreducible. We proceed by induction on m, assuming, without loss of generality, that m ≥ n. If m = 1, then s1 = r = t1 and the statement holds. Assume that m > 1. We have s1 s2 · · · sm = t1 t2 · · · tn , so sm | t1 t2 · · · tn . 3). Therefore, sm | tj for some j. By interchanging the factors tn and tj , if necessary, we may (and do) assume that sm | tn (for, if we prove the statement with this new ordering, then we can compose the permutation σ we get with the transposition (m, n) to get a permutation that works with the original ordering).

### Abstract Algebra II by Randall R. Holmes

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