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Unit 22: Polynomial Rings





          Thus, an example of an element from Z  [x] is f(x) =  2 x  +  3 x + i.                Notes
                                                       2
                                          4
          Here deg f(x) = 2, and the leading coefficient of f(x) is  2 .
          Now, for ring R, we would like to see if you can define operations on the set R [x] so that it
          becomes a ring. For this purpose we define the operations of addition and multiplication of
          polynomials.
          Definition: Let f(x) = a,, + a1x + .. + a, x  and g (x) = b  + b, x + .. + b x  be two polynomials in
                                                                   m
                                          n
                                                      0
                                                                 m
          R[X]. Let us assume that m 2 n. Then their sum f(x) + g(x) is given by
          f(x) + g(x) = (a,, + b ) + (a, + b )x + .. + (a + b,) x  + b  x  .. + b x .
                                                      n+l
                                                              m
                                 1
                                          n
                         0
                                                   n+1
                                               n
                                                            m
          For example, consider the two polynomials p(x), q(x).in Z[x] given by
          p(x) = 1 + 2x + 3x , q(x) = 4 + 5x + 7x 3
                        2
          Then
          p(x) + q(x) = (1+4) + (2+5)x + (3+0) x2 + 7x  = 5 + 7x + 3x  + 7x .
                                                           3
                                           3
                                                      2
          Note that p (x) + q (x)  Z [X] and that
          From the definition given above, it seems that deg (f(x)+g(x)) = max (deg f (x), deg g (x)). But this
          is not always the case. For example, consider p(x) = 1 + x  and q (x) = 2 + 3x – x  in Z [X].
                                                                         2
                                                       2
          Then p(x) + q(x) = (1+2) + (0+3)x + (1-1)x  = 3 a 3x.
                                          2
          Here deg (p(x) + q (x)) = 1 < max (deg p(x), deg q(x)).
          So, what we can say is that
          deg (f(x) + g(x))  max (deg f(x), deg g(x))
            f(x), g(x)  R[x].
          Now let us define the product of polynomials.
          Definition: If f(x) = a,, + a x + .. + a, x  and g(x) = b  + b, x + .. + b x  are two polynomials in R [x],
                                                              m
                                       n
                              1
                                                            m
                                                  0
          we define their product f(x). g(x) by
          f(x) . g(x) = c  a c x +.. + c m+n x m+n
                    0
                       1
          where c  = a b , + a b  + .... a b   i = 0,l ,... ; m + n.
                                  0 i
                         i-1  1
                    1 0
                1
          Note that a  = 0 for i > n and b  = 0 for i > m,
                                  i
                   i
          As an illustration, let us multiply the following polynomials in Z[x] :
          p(x) = 1 – x + 2x , q(x) = 2 + 5x + 7x .
                       3
                                      2
          Here a, = 1, a, = –1, a  = 0, a = 2, b  = 2, b, = 5, b  = 7.
                                                2
                           2
                                3
                                     0
                         5
                             i
          Thus, p(x) q(x) =    c x ,  where
                            i
                        i=0
          c  = a b  = 2,
           0
              0 0
          c  = a b  + a b  = 3,
           1
              1 0
                   0 1
          c  = a b  + a  b  + a b  = 2,
                         0 2
              2 0
                     1
                   l
           2
          c  = a b  + a b + a b + a b  = –3 (since b  = 0).
           3  3 0   2 1   1 2   0 3       3
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