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Let ²%³ ~  b  % b Ä b  % . Prove that if  is a prime for which  “  for  € ,  \“  Á  \“  then ²%³ is irreducible. 21. Show that if a polynomial  ²%³ £  b % is self-reciprocal and irreducible, then deg² ³ must be even. Hint: check the value of  ²c³. 22. Suppose that  ²%³ ~ ²%³²%³  - ´%µ, where ²%³ and ²%³ are irreducible, and  ²%³ is self-reciprocal. Show that either a) ²%³ ~ 9 ²%³ and ²%³ ~ 9 ²%³ with  ~ f , or b) ²%³ ~ 9 ²%³ and ²%³ ~ c 9 ²%³ for some   - .

Let ¸ ¹ be a basis for 3 as a vector space over - . Then  ÁÃÁ ~   ÁÃÁ ,   for  ÁÃÁ ,  - and so ²% Á à Á % ³ ~  @  ÁÃÁ ,  A% Ä%  ÁÃÁ  ~ @   ÁÃÁ , % Ä% A   ÁÃÁ Hence, the independence of the  's implies that the polynomial   ÁÃÁ , % Ä%  ÁÃÁ in - ´% Á à Á % µ is the zero function on - . As we have remarked, this implies that  ÁÃÁ , ~  for all  Á à Á  and  . … Common Divisors and Greatest Common Divisors In defining the greatest common divisor of two polynomials, it is customary (in order to obtain uniqueness) to require that it be monic.

Also, since the usual degree formula deg² ³ ~ deg² ³ b deg²³ holds when 9 is an integral domain, we get an immediate upper bound on the number of roots of a polynomial. … 28 Field Theory Note that if 9 is not an integral domain then the preceding result fails. For example, in { , the four elements Á Á and are roots of the polynomial % c  . From this, we get the following fundamental fact concerning finite multiplicative subgroups of a field. 4 Let - i be the multiplicative group of all nonzero elements of a field - .