By Randall R. Holmes

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**Example text**

For convenience, we will restrict our search to integral domains. The theorem can be thought of as saying that prime numbers are the building blocks for the integers greater than one. We need a generalization of these building blocks to our arbitrary integral domain. The reader most likely learned that a prime number is an integer greater than one having as positive factors only one and itself. This definition yields 2, 3, 5, 7, 11, and so on. The words “greater than” and “positive” here create obstacles for any generalization, since the axioms for an integral domain provide no notions of order or positivity.

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. 4 Subspace Let F be a field and let V be a vector space over F . A subspace of V is a subgroup of (V, +) that is closed under scalar multiplication. Thus, a subset W of V is a subspace if and only if 62 (i) 0 ∈ W , (ii) w, w ∈ W ⇒ w + w ∈ W , (iii) w ∈ W ⇒ −w ∈ W , (iv) a ∈ F, w ∈ W ⇒ aw ∈ W. We write W ≤ V to indicate that W is a subspace of V .

1 Theorem. If R is a UFD, then the polynomial ring R[x] is a UFD. Proof. ) Here are some applications of the theorem: • 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 with identity and let σ : R → R be a homomorphism.