**Proof.**
A regular local ring is Cohen-Macaulay, see Lemma 10.105.3. Hence the equivalences (1) $\Leftrightarrow $ (2) and (3) $\Leftrightarrow $ (4), see Proposition 10.102.4. By Lemma 10.105.4 the ideal $\mathop{\mathrm{Ker}}(A \to B)$ can be generated by $\dim (A) - \dim (B)$ elements. Hence we see that (4) implies (2).

It remains to show that (1) implies (4). We do this by induction on $\dim (A) - \dim (B)$. The case $\dim (A) - \dim (B) = 0$ is trivial. Assume $\dim (A) > \dim (B)$. Write $I = \mathop{\mathrm{Ker}}(A \to C)$ and $J = \mathop{\mathrm{Ker}}(A \to B)$. Note that $J \subset I$. Our assumption is that the minimal number of generators of $I$ is $\dim (A) - \dim (C)$. Let $\mathfrak m \subset A$ be the maximal ideal. Consider the maps

\[ J/ \mathfrak m J \to I / \mathfrak m I \to \mathfrak m /\mathfrak m^2 \]

By Lemma 10.105.4 and its proof the composition is injective. Take any element $x \in J$ which is not zero in $J /\mathfrak mJ$. By the above and Nakayama's lemma $x$ is an element of a minimal set of generators of $I$. Hence we may replace $A$ by $A/xA$ and $I$ by $I/xA$ which decreases both $\dim (A)$ and the minimal number of generators of $I$ by $1$. Thus we win.
$\square$

## Comments (3)

Comment #2968 by Dario Weißmann on

Comment #3093 by Johan on

Comment #3100 by Dario Weißmann on