Let $\mathcal{I}$ be a proper ideal of $\mathbb{F}[x]$ where $\mathbb{F}$ is a field. Show that there is a polynomial $P_0(x)\in \mathbb{F}[x]$ with degree larger than $0$ with the property that $$\mathcal{I}=\{P_0(x)\cdot q(x):q(x)\in \mathbb{F}[x]\}.$$ (In other words, prove there is a polynomial $P_0$ with the property that $\mathcal{I}$ consists precisely of the multiples of $P_0$.)
Previously, I have proved
Let $\mathbb{F}$ denote the field described the set $$\mathbb{F}=\{r+si: r,s\in \mathbb{Q}\}$$ (Here $i$ is the infamous element of $\mathbb{C}$ with $i^2=-1$.) Consider the function $\lambda: \mathbb{Q}[x] \to \mathbb{F}$ defined by $$\lambda(p)=p(i) \qquad \text{ for all } p\in \mathbb{Q}[x]$$
1) Show that $\lambda$ is a ring homomorphism. 2) Show that the kernel of $\lambda$ is the set of all polynomials $p\in \mathbb{Q}[x]$ that are divisible by $x^2+1$.
Proof of (i): For $\lambda$ to be a ring homomorphism, we need to show that > a) $\lambda(f+g)=\lambda(f)+\lambda(g)$ b) $\lambda(fg)=\lambda(f)\lambda(g)$ c) $\lambda(1)=1$
\textbf{\textit{Proof of a:}} Let $f,g \in \mathbb{Q}[x]$. Then we need to show that $\lambda(f+g)=\lambda(f)+\lambda(g)$. So $$\lambda(f+g)=(f+g)(i)=f(i)+g(i)=\lambda(f)+\lambda(g).$$
\textbf{\textit{Proof of b:}} Let $f,g \in \mathbb{Q}[x]$. Then we need to show that $\lambda(fg)=\lambda(f)\lambda(g)$. So $$\lambda(fg)=(fg)(i)=f(i)g(i)=\lambda(f)\lambda(g).$$
\textbf{\textit{Proof of c:}} Let $f=1 \in \mathbb{Q}[x]$. Then we need to show that $\lambda(1)=1$. So $$\lambda(1)=1(i)=1.$$
Hence $\lambda$ is a ring homomorphism.
Proof of (ii): We need to show that the kernel of $\lambda$ is the set of all polynomials $p\in > \mathbb{Q}[x]$ that are divisible by $x^2+1$.
Suppose $x^2+1|p(x)$. Then $p(x)=(x^2+1)g(x)$ for some $g(x)\in > \mathbb{Q}[x]$. Using part (i), we can show that $p(x) \in > ker(\lambda)$ (i.e. that $p(x)=0$): $$\lambda(p(x))=(x^2+1)g(x)=(i^2+1)g(i)=0\cdot g(i)=0$$ Hence we have show that $p(x)\in ker(\lambda)$.
Suppose $p(x) \in ker(\lambda)$. Using the Euclidean division over $\mathbb{Q}[x]$ to divide $p(x)$ by $x^2+1$. We obtain:
\begin{equation*} \begin{aligned} 0 & =\lambda(p(x)) \\ & > =\lambda(q(x)(x^2+1)+r(x)) \\ & =\lambda(q(x))\lambda(x^2+1)+\lambda(r(x)) \\ & =p(i)(i^2+1)+r(i) \\ & =r(i) \\ \end{aligned} \end{equation*} Hence we have show that the remainder is $0$, which implies that $p(x)=(x^2+1)g(x)$.
Hence, the kernel of $\lambda$ is the set of all polynomials $p\in \mathbb{Q}[x]$ that are divisible by $x^2+1$.
I know both proofs are related to one another. So what is needed to change in the previous proof for the new proof?
Let $\mathcal{I}$ be a proper ideal of $\mathbb{F}[x]$ where $\mathbb{F}$ is a field. We need to show that there is a polynomial $P_0$ with the property that $\mathcal{I}$ consists precisely of the multiples of $P_0$.
Suppose $A(x) \in \mathcal{I}$. Assume that $A(x)$ is irreducible. Using the Euclidean division over $\mathbb{F}[x]$ to divide $A(x)$ by $P_0(x)$. We obtain $A(x)=P_0(x)Q(x)+R(x)$ for some $R(x), Q(x)\in \mathbb{F}[x]$ where $0<\partial R<\partial P_0$. Let there be a $\alpha$ such that $A(\alpha)=0$. So
\begin{equation*} \begin{aligned} & 0=A(\alpha) \iff =P_0(\alpha)Q(\alpha)+R(\alpha) \\ \Rightarrow & P_0(\alpha)Q(\alpha)=-R(\alpha)=R'(\alpha) \\ \end{aligned} \end{equation*} This implies that $P_0$ divides $R'$ or $Q$ divides $R'$. So $\partial P_0\leq \partial R'=\partial R$ or $\partial Q \leq \partial R'=\partial R$ By this is a contradiction since $\partial R=\partial R'< \partial P_0$. Hence $A(x)=P_0(x)Q(x)$.
Hence $A(x)$ has satisfies the property that there is a polynomial $P_0$ such that $\mathcal{I}$ consists precisely of the multiples of $P_0$