Question

How to show that an element in a quotient ring of polynomial is not a zero element

Answer 1

Lemma. Let R be a commutative ring, and suppose $a(x), b(x), p(x) \in R[x]$ . Then $a(x)

  = b(x) \mod{p(x)}$ if and only if $a(x) +

  \langle p(x) \rangle = b(x) + \langle p(x) \rangle$ .

Proof. Suppose $a(x) = b(x)

  \mod{p(x)}$ . Then $a(x) = b(x) +

  k(x) \cdot p(x)$ for some $k(x) \in R[x]$ . Hence,

$$a(x) + \langle p(x) \rangle = b(x) + k(x) \cdot p(x) + \langle p(x) \rangle = b(x) + \langle p(x) \rangle.$$

Conversely, suppose $a(x) + \langle

  p(x) \rangle = b(x) + \langle p(x) \rangle$ . Then

$$a(x) \in a(x) + \langle p(x) \rangle = b(x) + \langle p(x) \rangle.$$

Hence,

$$a(x) = b(x) + k(x) \cdot p(x) \quad\hbox{for some}\quad k(x) \in R[x].$$

This means that $a(x) = b(x) \mod{p(x)}$ .

Depending on the situation, I may write $a(x) = b(x)

  \mod{p(x)}$ or $a(x) + \langle

  p(x) \rangle = b(x) + \langle p(x) \rangle$ .