Ideals and Quotient Rings -- Abstract Algebra 19 

Yes, this is a mistake - Michael got confused between the degree of a polynomial and the order of a polynomial, the latter being the _least_ power of x with a non-zero coefficient. So the ideal is I = {p(x) ϵ *Z* [x] : ord(p) ≥ 2}. For polynomials over any integral domain (here *Z* ) we have deg(pq) = deg(p) + deg(q), and for polynomials over a general ring we have deg(pq) ≤ deg(p) + deg(q) - likewise for the order we have over any integral domain that ord(pq) = ord(p) + ord(q), and over a general ring we have the inequality ord(pq) ≥ ord(p) + ord(q). There are also inequalities for deg(p+q) and ord(p+q). (these are inequalities even over integral domains; for example over *Z* we have deg(pq) ≤ max(deg(p), deg(q)) where equality may not occur, e.g. deg((x^2 + x) + (1 - x^2)) = deg(x + 1) = 1 < 2 = max(deg(x^2 + x), deg(1 - x^2)).)

No, quotient rings may be an integral domain. For example, if R is a ring and I is a prime ideal of R, then R/I is an integral domain

@@fakezpred gotcha 👍🏻 thank you!