Question

Let $\mathit{F}$ be a field. Prove that every ideal in $\mathit{F}\left[x\right]$is principal. [Hint: Use the Division Algorithm to show that the nonzero ideal $\mathit{I}$in $\mathit{F}\left[x\right]$is$\left(p\left(x\right)\right)$, where $\mathit{p}\left(x\right)$is a polynomial of smallest possible degree in I ].

Short answer

Hence, it is proved$I=\left(p\left(x\right)\right)$

Step-by-step solution

Consider the polynomial

Consider that $I$is an ideal in$F\left[x\right]$ .

Let us assume $\left(p\left(x\right)\right)\in I$ is the polynomial of the smallest possible degree in $I$.

Show that every ideal in F(x) is principal

Consider that $p\left(x\in I\right)$is the definition and I is an ideal, such that $\left(p\left(x\right)\right)\subset I$.

On the other hand, consider$f\left(x\right)\in I$.

Apply polynomial division algorithm as

$q\left(x\right),r\left(x\right)\in F\left[x\right]$, such that$q\left(x\right)p\left(x\right)+r\left(x\right)$.

So, either $r\left(x\right)=0$ or$deg\left(r\left(x\right)\right)<deg\left(p\left(x\right)\right)$.

Therefore, $r\left(x\right)=f\left(x\right)-q\left(x\right)p\left(x\right)\in I$.

Since $f\left(x\right),p\left(x\right)\in I,r\left(x\right)\ne 0$, contradicts the slight assumption on $p\left(x\right)$,

Hence, $r\left(x\right)=0$.Thus,$f\left(x\right)\in \left(p\left(x\right)\right)$.

Hence, it is proved $I=\left(p\left(x\right)\right)$.