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Example text

For each 1 ≤ k ≤ n, the set {(0, . . 0, a, 0, . . , 0) | a ∈ F } k is a subspace of F n . • Let n be a nonnegative integer. The set of all polynomials over F of degree ≤ n is a subspace of F [x]. • Let n be a positive integer. The set {[aij ] | aij = 0 for all i > j} of all upper triangular n×n matrices over F is a subspace of Matn (F ). 5 Quotient space Let V be a vector space and let W be a subspace of V . Then W is a subgroup of the additive group (V, + ) and, since this latter group is abelian, W is normal.

49 (i) The element r of R is a zero of f (x) if and only if x − r divides f (x). (ii) If f (x) is nonzero, then it has at most deg f (x) zeros in R. Proof. (i) Let r ∈ R. Assume that r is a zero of f (x). 3), there exist polynomials q(x) and s(x) over R with deg s(x) < deg(x−r) = 1 such that f (x) = q(x)(x−r)+s(x). Now deg s(x) is 0 or −∞, so in either case s(x) is constant. Since 0 = f (r) = q(r)(r − r) + s(r) = s(r), it follows that s(x) is the zero polynomial. Therefore, f (x) = q(x)(x − r), and x − r divides f (x).

For instance, x2 + 1 is irreducible over R, but over C it factors as (x + i)(x − i). Nonetheless, it is the case that if f (x) is irreducible over Z, then it is irreducible over Q as well. This fact is stated in the theorem below. In order to prove the theorem we need a definition and a lemma. The polynomial f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ Z[x] is primitive if its coefficients are relatively prime, that is, if gcd(a0 , a1 , a2 , . . , an ) = 1. 1 Lemma (Gauss). A product of primitive polynomials is primitive.

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Abstract Algebra II by Randall R. Holmes


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