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Sec 45 #26. Prove that if p is an irreducible in a UFD, then p is a prime. Hint. Let D be a UFD, and let p be an irreducible in D. Let a, b € D such that plab .We need to show that pla or plb. If ab=0, then either a=0 and b=0 (why). We are done in this case (why). Suppose ab # 0. Then neither a nor b is zero (otherwise what happens?). As plab, write ab = cp for some c in D. Then there again arise two cases: Case 1) Either a or b is a unit: If a is a unit, we have a-¹ (ab) = a-¹ (cp) so that b = and proceed. If b is a unit, argue similarly. Case 2) Neither a nor b is a unit: From ab=cp, show that C is nonzero and nonunit in D. Thus a, b and C will be nonzero and nonunits in D. Since D is a UFD, so a, b and C have factorizations a=P1 Pm, b=q1- In and c=r1 rk into irreducibles Pi, aj and r1 in D. Use these factorizations into ab=cp and as the two sides the resulting equation are equal, their factorizations into irreducibles must be unique (up to associates). Then either p and some pi are associates, or p and some aj are associates. Then either = (why), or = (why). But a=P1 Pm yields a E< Pi >, and Thus a E< > or and proceed to conclusion.

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