By Randall R. Holmes

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**Additional info for Abstract Algebra II**

**Sample text**

But then (s) ⊆ (ri ) (ri+1 ) ⊆ (s), a contradiction. 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).

In high school algebra, the polynomial f (x) = ni=0 ai xi is really a function f : R → R, which is not a vague concept. If we attempted to define the polynomial f (x) = ni=0 ai xi ∈ R[x] as the n i function R → R sending x to i=0 ai x we would lose something. For instance, if R = Z2 , then the polynomials x and x2 are not equal (since their corresponding coefficients are not equal), but as functions Z2 → Z2 they are equal since they both send 0 → 0 and 1 → 1. 2 Degree of polynomial Let R be an integral domain.

Here are some applications of the theorem: 52 • Since Z is a UFD, so is Z[x]. • A field F is a UFD, so F [x] is a UFD as well. 6). • If R is a UFD, then so is the ring of polynomials over R in n indeterminants R[x1 , x2 , . . , xn ] (n ∈ N) defined recursively by putting R[x1 , x2 , . . , xn ] ∼ = R[x1 , x2 , . . , xn−1 ][xn ]. This claim follows immediately from the theorem by using induction on n. 8 Induced homomorphism of polynomial rings Let R and R be commutative rings and let σ : R → R be a homomorphism.

### Abstract Algebra II by Randall R. Holmes

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