Every ideal in F[x] is principal

Dependencies:

1. Principal ideal
2. Polynomial division theorem
3. p(x)F[x] = F[x] iff p is a non-zero constant

Let $F$ be a field. Every ideal of $F[x]$ is a principal ideal, i.e. of the form $p(x)F[x]$.

Proof

A trivial ideal is a principal ideal, because $\{0\} = 0F[x]$ and $F[x] = 1F[x]$.

Let $I$ be a non-trivial ideal of $F[x]$. Let $p(x)$ be the polynomial of least degree in $I - \{0\}$. Since $I$ is an ideal, $p(x)F[x] \subseteq I \subseteq F[x]$.

When $p \in F-\{0\}$, $p(x)F[x] = F[x] = I$. So $I$ is a principal ideal.

Let $p(x) \not\in F$. Let $a(x) \in I$ such that $p \not\mid a$ in $F[x]$. Then by the polynomial division theorem, $\exists q, r \in F[x]$ such that $a = pq + r$ and $\deg(r) < \deg(p)$ and $r \neq 0$.

\begin{align} & a, p \in I \\ &\Rightarrow a, pq \in I \tag{$\because$ $I$ is a field} \\ &\Rightarrow r = a - pq \in I \tag{closure in ring $I$} \end{align}

Since $r \in I - \{0\}$, $\deg(r) < \deg(p)$ and $p$ is a least degree polynomial in $I - \{0\}$, we have a contradiction. Therefore, $p \mid a$ for all $a(x) \in I$. Therefore, $I = p(x)F[x]$.

Dependency for: None

Info:

• Depth: 6
• Number of transitive dependencies: 20

Transitive dependencies:

1. Group
2. Identity of a group is unique
3. Subgroup
4. Inverse of a group element is unique
5. Conditions for a subset to be a subgroup
6. Condition for a subset to be a subgroup
7. Ring
8. Polynomial
9. Degree of sum of polynomials
10. Degree of product of polynomials
11. Zero divisors of a polynomial
12. Conditions for a subset of a ring to be a subring
13. Ideal
14. Principal ideal
15. Field
16. 0x = 0 = x0
17. Integral Domain
18. A field is an integral domain
19. Polynomial division theorem
20. p(x)F[x] = F[x] iff p is a non-zero constant