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In an integral domain deg(pq) = deg(p) + deg(q)

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##R## be an integral domain and ##p_1(x),p_2(x) \in R[x]## with neither equal to ##0##. Show that the degree of ##p_1(x)p_2(x)## is the sum of the degrees of ##p_1(x)## and ##p_2(x)##.

    2. Relevant equations


    3. The attempt at a solution
    Here is my attempt.
    Let ##p_1(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0## and ##p_2(x) = b_m x^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0##. Evidently ##\operatorname{deg} (p_1) = n## and ##\operatorname{deg} (p_2) = m##

    Upon multiplication of ##p_1## and ##p_2##, the highest order term will be ##a_n b_m x^{n+m}##. Since ##a_n, b_m \in R##, an integral domain, and ##a_n \ne 0## and ##b_m \ne 0##, it must be the case that ##a_n b_m \ne 0##. Hence ##\operatorname{deg} (p_1 p_2) = \operatorname{deg} (p_1) + \operatorname{deg} (p_2) = m + n##.
     
  2. jcsd
  3. Apr 5, 2017 #2

    fresh_42

    Staff: Mentor

    I would have worded it differently (##a_n \neq 0 \neq b_m## at the beginning as part of the definition of degree), but it is correct.
     
  4. Apr 5, 2017 #3
    Also, one thing. To what extent is it rigorous enough to just say that "Upon multiplication of ##p_1## and ##p_2##, the highest order term will be ##a_n b_m x^{n+m}##"?
     
  5. Apr 6, 2017 #4

    fresh_42

    Staff: Mentor

    It is. You can write ##p_1(x)\cdot p_2(x) = (a_nx^n + \ldots )\cdot (b_mx^m + \ldots ) = a_nb_mx^{n+m} + \ldots ## or even
    $$
    p_1(x)\cdot p_2(x) = \left( \sum_{i=0}^{n}a_i x^i \right) \cdot \left( \sum_{j=0}^{m}b_j x^j \right) = \sum_{k=0}^{n+m} \sum_{i+j=k}a_ib_j x^{i+j}
    $$
    instead, but this isn't really necessary in this case.
     
  6. Apr 6, 2017 #5
    An integral domain contains no zero divisors, therefore leading term will not vanish as you correctly deduced.
     
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