How to find the generator of the following ideals
$\cal a=${$F\in \mathbb Q[X]:F(i)=0$} in $\mathbb Q[x]$,
$\cal b=${$F\in \mathbb Q[X]:F(\sqrt 2i)=0$} in $\mathbb Q[x]$?
$\cal c=${$F\in \mathbb R[X]:F(\sqrt 2)=0$} in $\mathbb R[x]$,
$\cal d=${$F\in \mathbb R[X]:F(i\sqrt 2)=0$} in $\mathbb R[x]$.
All I know is that I have to find the polynomial of least degree $F(x)\in \mathbb Q[x]$ such that $F(x)=0.$
Solution(1)
By inspection, $F(x)=x^2+1$. So, $\cal a=\langle x^2+1\rangle$
Solution(2)
By inspection, $F(x)=x^2+2$. So, $\cal b=\langle x^2+2\rangle$
Solution(3)
By inspection, $F(x)=x^2-2$. So, $\cal c=\langle x-\sqrt 2\rangle$
Solution(4)
By inspection, $F(x)=x^2+2$. So, $\cal d=\langle x^2+2\rangle$
Are the above solutions correct? Is there any analytic method of obtaining the generator of ideals?
Welcome to MSE! Here is an algebraic explanation.
Consider an ideal $I$ in $K[x]$, where $K$ is a field. Since $K[x]$ is a principal ideal domain, there is a polynomial $f\in K[x]$ such that $I=\langle f\rangle$.
Consider the quotient ring $K[x]/\langle f\rangle$. The polynomial $f$ has a zero in the quotient ring, namely the residue class $\bar x = x+\langle f\rangle$, since $f(\bar x) = 0$ in the residue class ring. The polynomial $f$ has smallest degree with the property $f(\overline x)=0$ and so is irreducible over $K$, i.e., the minimum polynomial of $\bar x$ over $K$.
So what you are looking for is the minimum polynomial $f\in K[x]$ of an element $a\in L$ in an extension field $L$ over $K$.