By Randall R. Holmes

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**Extra info for Abstract Algebra II**

**Sample text**

C) Let S ≤ R. The set ϕ(S) = {ϕ(s) | s ∈ S} is known to be a subgroup of (R , +) by group theory. If s, t ∈ S, then st ∈ S and ϕ(s)ϕ(t) = ϕ(st), so ϕ(S) is closed under multiplication. Therefore, it is a subring of R . (d) Let S ≤ R . The set ϕ−1 (S ) = {r ∈ R | ϕ(r) ∈ S } is known to be a subgroup of (R, +) by group theory. If r, s ∈ ϕ−1 (S ), then ϕ(rs) = ϕ(r)ϕ(s) ∈ S , which implies that rs ∈ ϕ−1 (S ), so ϕ−1 (S ) is closed under multiplication. Therefore, it is a subring of R. 4 Kernel and image Let ϕ : R → R be a homomorphism of rings.

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 42 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).

The function ϕ is given by ϕ(a + I) = ϕ(a). Proof. Let I be an ideal of R with I ⊆ ker ϕ. As in the statement of the theorem, let ϕ : R/I → R be the function given by ϕ(a + I) = ϕ(a). If a + I = b + I (a, b ∈ R), then a − b ∈ I ⊆ ker ϕ, so that ϕ(a) − ϕ(b) = ϕ(a − b) = 0, implying ϕ(a) = ϕ(b). Thus, ϕ is well defined. For a + I, b + I ∈ R/I, we have ϕ((a + I) + (b + I)) = ϕ((a + b) + I) = ϕ(a + b) = ϕ(a) + ϕ(b) = ϕ(a + I) + ϕ(b + I) and similarly ϕ((a + I)(b + I)) = ϕ(a + I)ϕ(b + I), so ϕ is a homomorphism.

### Abstract Algebra II by Randall R. Holmes

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