Let $I=\langle 4, 2x,x^{2} \rangle$ in $\mathbb{Z[x]}$. Show that $I$ can not be generated by 2 elements.

Question

I was trying like this: Let $I=\langle f(x),g(x) \rangle$. Then $4=f(x)q_{1}(x)+g(x)q_{2}(x)$. But $deg (4)=0$. So $deg(f(x))=0$ and $deg(g(x))=0$. So $f(x)=a$ and $g(x)=b$ where $a,b \in \mathbb{Z}$. Then I am stuck. Is this the correct way? Any help is appreciated.

Answer 1

Suppose one has a commutative unital ring $R$ with an ideal $I$, and a maximal ideal $M$. Then $R/M$ is a field $k$, say. Also $I/MI$ is a module over $R/M$, a vector space over $k$. Suppose its dimension as a vector space is $d$. If $a_1,\ldots,a_n$ are generators of $I$, then their images in $I/MI$ span $I/MI$ as a $k$-vector space. Therefore $n\ge d$.

Here, $R=\Bbb Z[X]$, $I=\left<x^2,2x,4\right>$, $M=\left<x,2\right>$ and $d=3$.