Generator for Ideal

The set that will allow for creating an Ideal. Often you use Polynomial Long Division to find the generator

Definition

• For Ring $(R,+,x)$
• For Ideal $I\subset R$
• The generator $G$ is the subset of elements $G\subset I$ that allows for all elements of $G$ to be represented as sums of multiples of $g_i\in G$ for elements of $r_i\in R$ ($\forall i\in I,\exists g_1,\dots ,g_k\in G,\exists r_1,\dots ,r_n\in R\text{ s.t }i={\sum }_{i=0}^n{r}_i{g}_i$)

For Principle Ideal, the set $G$ only has one element.

Finding Singleton

For a given ideal generated by: $⟨a_1,a_2,\dots ⟩$, we get that the singleton is: $gcd(a_1,a_2,\dots )$

Example

Problem

With ${\mathbb{Z}}_7[x]$, show the set of polynomials annihilated at $x=4$ is an ideal.

Soln

Showing non-empty:

• Suppose $g\in I$ and $f\in {\mathbb{Z}}_7[x]$. (We suppose, because we dont know these exist yet)
• Then, $(fg)(4)=f(4)g(4)=f(4)(0)=0$
• Thus, $fg\in I$
• Consider $g(x)=x+3$
• Then, $g(4)=4+3=7\mathrm{mod}7=0$
• Thus, $g\in I$ Then, it follows that $I$ is non-empty!

Problem

Show that the set of $x+3$ is the generator

Soln

• Suppose $f\in I$, then by defn, $f(4)=0$
• Then, by the Fundamental Remainder Theorem, there exists a $g\in {\mathbb{Z}}_7[x]$ s.t $f(x)=(x+(-4))\ast g(x)$
• $=(x+3)\ast g(x)$ as $3$ is the Additive Inverse of $4$ in ${\mathbb{Z}}_7$
• Thus, $\exists g\in \mathbb{R}$ s.t $f=(x+3)\ast g$
• As $f$ is a generic element, the set $x+3$ is a generator for $I$