Reducing the degree of a polynomial

Question

Problem: Let R be an Integral domain. Let p $\in$ R[x]. Suppose deg(p) > 0. Show that multiplying p by other polynomials cannot bring down the degree.

Attempt: This makes good intuitive sense because if I would want to reduce the degree then I would let R be something like $Z/6Z$, but that is not an ID. I figure there must exist some proof by way of contradicting on the zero-divisor condition of Integral Domains but I cannot seem to find it.

Answer 1

Let $p(x) = a_nx^n + a_{n-1}x^{n-1} \cdots +a_0$ where $a_n \neq 0$. If $q(x) = b_m x^m + b_{m-1}x^{m-1}+ \cdots + b_0$ is another polynomial, where $b_m \neq 0$, then, when we multiply $p$ and $q$, we obtain

$$p(x)q(x) = a_n b_m x^{m+n} + (a_{n-1}b_m + a_n b_{m-1})x^{n+m-1}+ \cdots + a_0 b_0.$$

Since $R$ is an integral domain we have $a_n b_n \neq 0$ because $a_n \neq 0$ and $b_m \neq 0$. Therefore, $\deg(pq)= n+m \geq \deg(p)$.

Answer 2

Take the lead terms of two polynomials $a_m x^m$ and $b^n x^n$. Since $R$ is an integral domain, $a_m b_n \not = 0$. The $x^{m+n}$ term of the product of these polynomials is $a_mb_n x^{m+n}$.