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:<math>\phi(b(x)^m) = (b_1^m(x) + \langle p_1(x) \rangle, \ldots, b^m_s(x) + \langle p_s(x) \rangle).</math>
Now, each <math>b_i(x) + \langle p_i(x)\rangle</math> is an element of a field of order <math>q^d</math>, as noted earlier. The multiplicative subgroup of this field has order <math>q^d-1</math> and so, unless <math>b_i(x)=0</math>, we have <math>b_i(x)^{q^d-1}=1</math> for each ''i'' and hence <math>b_i(x)^m = \pm 1</math> for each ''i''. If <math>b_i(x)=0</math>, then of course <math>b_i(x)^m=0</math>. Hence <math>b(x)^m</math> is a polynomial of the same type as <math>a(x)</math> above. Further, since <math>b(x) \neq 0, \pm1</math>, at least two of the sets <math>A,B</math> and ''C'' are non-empty and by computing the above GCDs we may obtain non-trivial factors. Since the ring of polynomials over a field is
==Applications==
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