1. ## Prove that deg(f(x)g(x))=deg(f(x))+deg(g(x))

a.) Let R be an integral domain and f(x), g(x) are elements of R[x]. Prove that deg(f(x)g(x))=deg(f(x))+deg(g(x)).

b.) Is part (a) true if R is not an integral domain?

2. Originally Posted by Slazenger3

a.) Let R be an integral domain and f(x), g(x) are elements of R[x]. Prove that deg(f(x)g(x))=deg(f(x))+deg(g(x)).

b.) Is part (a) true if R is not an integral domain?
a) let $f(x)=\sum_{i=0}^n a_ix^i, \ g(x)=\sum_{i=0}^m b_ix^i, \ a_n \neq 0, \ b_m \neq 0.$ then the leading term of $f(x)g(x)$ is $a_nb_mx^{n+m}$ and $a_nb_m \neq 0,$ because $a_n \neq 0, \ b_m \neq 0$ and $R$ is an integral domain.

b) no! a counter-example: $R=\mathbb{Z}/4\mathbb{Z}, \ f(x)=g(x)=\bar{2}x.$

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### deg(f(x)g(x))=deg(f(x)) deg(g(x))

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