Abstract Algebra by Prabhat Choudhary

By Prabhat Choudhary

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A polynomial that is not irreducible is said to be reducible orJactorable. For example, X2 + 1, regarded as an element of lR [X], where lR is the field of real numbers, is irreducible, but if we replace R by the larger field C of complex numbers, X2 + 1 is factorable as (X - i)(X + i), i irreducible over C. = J=l. We say that X2 + 1 is irreducible over lR but not Now consider D[X], where D is a unique factorization domain but not necessarily a field, for example, D = Z . The polynomial 12X+ 18 is not an irreducible element of Z [X] because it can be factored as the product of the two nonunits 6 and 2X + 3.

The leading coefficient of g is 1. For example (with R = Z), we can divide 2X3 + 1OX2 + 16X +10 by X2 +3X +5: 2X3 + IOX2 + l6X + 10 = 2X(X2 + 3X + 5) + 4X2 + 6X + 10. The remainder 4X2 + 6X + 10 does not have degree less than 2, so we divide it by X2 + 3X+ 5: 4X2 + 6X"'+ 10 = 4(X2 + 3X + 5) - 6X - 10. Combining the two calculations, we have • 2X3 + IOX2 + 16X + 10 = (2X + 4 )(X2 + 3X + 5) + (- 6X - 10) which is the desired decomposition. Ring 19 Division Algorithm If/and g are polynomials in R[X], with g monic,there are unique polynomials q and r in R[X] such that / = qg + rand deg r < deg g.

Show is a prime ideal if and only if n is a prime number. 2. Let I be a nontrivial prime ideal in the ring of integers. Show that in fact I must be maximal. 3. Let F[[X]] be the ring of formal power series with coefficients in the field F. Show that (X) is a maximal ideal. Ring 18 4. Let I be a proper ideal of F[[X]]. Show that I ~ (X), so that (X) is the unique maximal ideal of F[[X]]. ) 5. Show that every ideal of F[[X]] is principal,and specifically every nonzero ideal is of the form (xn) for some n = 0, 1, ....

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