By Prabhat Choudhary

ISBN-10: 1441651799

ISBN-13: 9781441651792

ISBN-10: 8189473549

ISBN-13: 9788189473549

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**Additional info for Abstract Algebra**

**Example text**

Suppose thatJis a nonzero polynomial in D[X] and thatJ can be factored as gh, where g and h belong to F[X]. Then there is a nonzero element).. g E D[X] and).. -Ih E D[X]. Thus ifJis factorable over F, then it is factorable over D. Equivalently, ifJis irreducible over D, then Jis irreducible over F. Proof The coefficients of g and h are quotients of elements of D. If a is the least common denominator for g (technically, the least common multiple of the denominators of the coefficients of g), let g* = ag E D[X].

If f is a nonconstant polynomial over the field F, and f has no roots in F, we can always produce a root offin an extension field of F. We do this after a preliminary result. ' Thenfis a monomorphism. ' Field 39 Proof First note that afield F has no ideals except {O} and F. For if a is a nonzero member of the ideal I, then ab = 1 for some b E F, hence 1 E I, and therefore I = F. Taking I to be the kernel of J, we see that I cannot be all of F because j{l) 7: O. Thus I must be {O}, so thatfis injective.

Now let g*(X) = go + glX + ... + g~, h*(X) = ho + hlX + ... + h~. Since p is a prime factor of c = ab and abJ= g*h*, p must divide all coefficients of g*h*. If P does not divide every gi and p does not divide every hi' let gu and hv be the coefficients of minimum index not divisible by p. Then the coefficient of XU+ v in g*h* is gohu+v + glh u+v_1 + ... + guhv + ... + gu+v-Ihl + gu+vho' Ring 35 But by choice of u and v, p divides every term of this expression except guhv' so that p cannot divide the entire expression.

### Abstract Algebra by Prabhat Choudhary

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