My textbook says: Let $A$: Local ring with the unique maximal ideal $m$, $d:=\dim A$ then $f_1, f_2,\ldots,f_d \in m$, $m=\sqrt{(f_1,\ldots,f_d)} \iff \dim A/(f_1,...f_d)=0$.
How can I prove this? What I could understand was $\dim R=0$ implies that any prime ideal of $R$ is also a maximal ideal....
Let $(R,\mathfrak{m})$ be a local ring and $I$ be an ideal such that $\sqrt{I}=\mathfrak{m}$. The ideal $I$ is known as a parameter ideal for $R$.
$\Rightarrow$: We have $\dim R/I=0$. Indeed, it is clear that $\dim R/I=\dim R/\mathfrak{m}=0$ since $R/\mathfrak{m}$ is a field and by 1.
$\Leftarrow$: Let $J\subseteq\mathfrak{m}$ be an ideal such that $\dim R/J=0$. Every minimal prime in $R/J$ is maximal. Since $R$ is local, it follows that $\mathfrak{m}/J$ is the unique minimal prime ideal of $R/J$. Hence, we have $\sqrt{J}=\mathfrak{m}$ and $J$ is a parameter ideal for $R$.
1. I repeatedly used the fact that for a ring $R$ and an ideal $I$ in $R$, we have $$ \sqrt{I}=\bigcap_{\mathfrak{p}\colon\text{ prime containing }I}\mathfrak{p}=\bigcap_{\mathfrak{p}\colon\text{ minimal prime containing }I}\mathfrak{p}\,. $$
2. If $R$ is a local noetherian ring, then the minimal number of generators of a parameter ideal is equal to the dimension of $R$. If we choose $\dim R$ generators of a parameter ideal, they are called a system of parameters for $R$.