$I^{j}/I^{j+1} \cong R/I$ for any ideal I in ring R.

Question

Let $R$ be a commutative ring with $1$ and $I$ be an ideal in it. Let $\overline{\alpha} \in I^j\setminus I^{j+1}$ and define $\theta\colon R \to I^j/I^{j+1}$ by $\theta(x)=\overline{\alpha x}$. My question is whether $I=\ker(\theta)$.

Answer 1

Take the ring $R = k[x,y]$, $k$ a field. Take the maximal ideal $I = (x,y)$. Then $R/I \cong k$ and so it is a one dimensional $k$-vector space. But $I^n / I^{n+1}$ is an ${n+1 \choose n}$-dimensional $k$ vector space.

Note that the number of monomials of degree $n$ in $D$ variables is ${n+D-1 \choose n}$.

Answer 2

No! This is not true. $R$ may contains zero-divisors and the element $\alpha$ may be an zero-divisor with the property that $\alpha \beta = 0, \beta \notin I.$

Example: $R = \mathbb Z_4[x], I=(x).$ Then $I^2=(x^2).$ Choose $\alpha = 2x.$ Then $2 \in \ker \theta.$ But $2 \notin I.$