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

Let $p(x)\in F[x]-F\Rightarrow \mathrm{deg}(p)\ge 1$. Since $F$ has no zero-divisors, all non-zero elements in $p(x)F[x]$ have degree at least 1. Therefore, $1\notin p(x)F[x]$ but $1\in F[x]$. Therefore, $p(x)F[x]\ne F[x]$.

Dependency for:

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

Info:

• Depth: 4
• Number of transitive dependencies: 9

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