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I wonder the smallest cardinality of generating set of maximal ideal in a ring $R$ in it coincide with the Krull dimension? i.e. Let $J=\{$Maximal ideal in $R\}$, $S_{I}$ is the generating set of maximal ideal $I$,Krull dimension of $R=n$. Is $min_{I\in J}|S_{I}|=n$?

If $R$ is polynomial ring or $C([0,1])$ the cases are true, but can we generalize to any commutative $R$?

Ken.Wong
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No. For any commutative ring $R$ and $I\leqslant R$ any ideal, denote by $\mu(I)$ the cardinality of a smallest generating set for $I$. If $R$ is Noetherian, then the best we can do in general is Krull's height theorem, which tells us that, for any $P\geqslant I$ a minimal prime ideal over $I$, we have $\text{ht}(P)\leqslant\mu(I)$. However, equality certainly does not have to hold. Let $F$ be your favorite field, and consider the ring $R:=F[x_1,\dots,x_n]\big/(x_ix_j)_{1\leqslant i,j\leqslant n}$. Then $R$ is local with $\dim R=0$ (why?), but the unique maximal ideal of $R$, $(\overline{x_1},\dots,\overline{x_n})$, is not generated by fewer than $n$ elements (why?). You can extend this example to the ring $F[x_i]_{i\in\mathbb{N}}\big/(x_ix_j)_{i,j\in\mathbb{N}}$, which is still local of Krull dimension $0$, but doesn't even have a finitely generated maximal ideal.

Atticus Stonestrom
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    Sorry if I'm overlooking something, but why is the dimension of $R$ equal to zero? For example, if $R=k[x,y]/(xy)$, the dimension of $R$ is $1$, right? ($R$ is the union of two lines, so it is natural to expect that it has dimension $1$.) – cqfd Jan 16 '21 at 15:39
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    @ShiveringSoldier ah, so in this case the indices are ranging over all possible values of $i$ and $j$, not just those where $i\neq j$... so eg when $n=2$, the ring is $F[x,y]\big/(x^2,xy,y^2)$ rather than $F[x,y]\big/(xy)$ – Atticus Stonestrom Jan 16 '21 at 16:09