p(x)F[x] = F[x] iff p is a non-zero constant

Dependencies:

  1. Ideal
  2. Degree of sum of polynomials
  3. A field is an integral domain

Let $F$ be a field. $p(x)F[x] = F[x]$ iff $p \in F-\{0\}$.

Proof

$p(x)F[x] \subseteq F[x]$.

Let $p \in F-\{0\}$ and $f(x) \in F[x]$. Then $f(x) = p(p^{-1}f(x)) \in pF[x] \Rightarrow F[x] \subseteq pF[x]$. Therefore, $pF[x] = F[x]$.

If $p(x) = 0$, $p(x)F[x] = \{0\} \neq F[x]$.

Let $p(x) \in F[x] - F \Rightarrow \deg(p) \ge 1$. Since $F$ has no zero-divisors, all non-zero elements in $p(x)F[x]$ have degree at least 1. Therefore, $1 \not\in p(x)F[x]$ but $1 \in F[x]$. Therefore, $p(x)F[x] \neq F[x]$.

Dependency for:

  1. The ideal generated by an irreducible polynomial is maximal
  2. Every ideal in F[x] is principal

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Polynomial
  4. Degree of sum of polynomials
  5. Integral Domain
  6. 0x = 0 = x0
  7. Ideal
  8. Field
  9. A field is an integral domain