At the maximal <ideal> $\mathfrak m=(2,X)$, the images of $4,2X,X^2$ form a three-dimensional <basis> of $I/\mathfrak mI$ over $\mathbb F_2$, so two generators are impossible. More generally $(2^{n-1},2^{n-2}X,\ldots,2X^{n-2},X^{n-1})$ needs exactly $n$ generators by the same argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 1E / b / Solution

 Solution (source code)