In an Artinian local ring, the maximal ideal is the set of nilpotent elements. In particular if \(x \in \mathfrak {m}_A\), then \(f (x)\) is not a unit.

We prove a more general result: if \(M\) is an \(A\)-module, then \(\operatorname{length}_\Lambda (M) = \operatorname{length}_A (M)\). This result follows easily by induction on \(\operatorname{length}_\Lambda (M)\) from the case \(\operatorname{length}_\Lambda (M) = 1\), in which case we have \(M \cong k\).

[

Let \(f : X \to Z\) and \(g : Y \to Z\) be morphisms in \(\mathbf{C}_\Lambda \). Consider the set \(P = \{ (x, y) \in X \times Y ~ \vert ~ f (x) = g (y)\} \). Since \(f\) and \(g\) are local, the subset \(\mathfrak {m}_X \times \mathfrak {m}_Y\) is an ideal (which is maximal). We have \(\mathscr {k}_\Lambda \to \mathscr {k}_P \to \mathscr {k}_X\), so \(\mathscr {k}_P \cong k\). Furthermore, since \(X\) and \(Y\) are Artinian \(\Lambda \)-modules by lemma 3, \(P\) is an Artinian \(\Lambda \)-module as well, so it is an Artinian ring.

Since \(p\) is surjective, we have \(\operatorname{length}_\Lambda (B) = \operatorname{length}_\Lambda (A) + \operatorname{length}_\Lambda (\ker p)\), and we also have that \(\operatorname{length}_B (\ker p) = 1\) if and only if \(\ker p\) is a minimal non-zero ideal, so using lemma 3 we get \((2) \iff (3)\). We now show \((1) \iff (2)\).

Assume that \(p\) is a small extension, write \(\ker p = (t)\), let \(I {\lt} \ker p\) be an ideal and \(x \in I\). There exists \(y \in B\) such that \(x = ty\). Since \(I \ne (t)\), \(y\) is not a unit, so we have \(y \in \mathfrak {m}_B\), and \(x = 0\) by definition of a small extension. So \(I=0\) and \(\ker p\) is minimal.

Assume that \(\ker p\) is a minimal non-zero ideal. Let \(t \in \ker p \setminus \{ 0\} \). By minimality, we have \(\ker p = (t)\). Furthermore, since \(\mathfrak {m}_B\) is nilpotent and \(\ker p \neq 0\) we have \(\mathfrak {m}_B \cdot \ker p \neq \ker p\) so \(\mathfrak {m}_B \cdot \ker p = 0\) by minimality. So \(p\) is a small extension.

By induction on \(\operatorname{length}_\Lambda (B) \in \mathbb {N}\). If \(\ker p = 0\), then \(p\) is an isomorphism. Otherwise, there exists a minimal non-zero ideal \(I\) of \(B\) because \(B\) is Artinian. By minimality, since \(\ker p \neq 0\), we have \(I \le \ker p\), so \(p\) factorizes through the small extension \(B \to B/I\). We conclude by applying the induction hypothesis to the map \(B/I \to A\).

One way is to apply the previous lemma to the map \(p : B \to k\) (i.e. with the property on morphisms \(q : X \to Y\) given by if \(Y = k\) then \(C (X)\)). In the code the proof is essentially copied and adapted from the previous lemma.

An Artinian ring is Noetherian, and its \(\mathfrak {m}\)-adic topology is discrete, so it is trivially complete.

A Noetherian ring is Artinian if and only if it has Krull dimension 0, so it suffices to show that any prime ideal is maximal. Let \(I\) be a prime ideal. Since \(\mathfrak {m}^n = 0 \le I\) we have \(\mathfrak {m}\le I\), so \(I = \mathfrak {m}\) is maximal.

\(A / \mathfrak {m}^n\) is a local Noetherian ring, with nilpotent maximal ideal \(\mathfrak {m}/ \mathfrak {m}^n\) and residue field \(k\), so lemma 12 applies.

There exists \(n\) such that \(\mathfrak {m}_A^n = 0\), so \(\mathfrak {m}_R^n \le u^{-1} (\mathfrak {m}_A ^n) = \ker u\).

This is a special case of the following lemma [ :

Let \(M \in \operatorname{Mod}_R^{fg}\). Write \(s : M \times M \to M\) for the addition map, \(\lambda _r : M \to M\) for the scalar multiplication by \(r \in R\), and \(z : 0 \to M\). The module structure on \(L(M)\) is defined using the images of these morphisms by \(L\), together with the identification \(L(M\times M) \cong L(M) \times L(M)\), and the fact that \(L(0)\) is terminal in \(Sets\) so it is a singleton. Checking that this defines a module structure is rather straightforward, by writing suitable commutative diagrams in \(\operatorname{Mod}_R^{fg}\) and pushing them with \(L\).

This is a very simple check.

By induction on the dimension.

If \(A\) is in \(\mathbf{C}_\Lambda \), there exists \(N \in \mathbb {N}^*\) such that \(\mathfrak {m}_A^N=0\), so we have \(F(A/\mathfrak {m}_A^n) \cong F(A)\) for all \(n \ge N\).

This is essentially the naturality of \(\hat{F}|_{\mathbf{C}_\Lambda } \cong F\).

In fact, we have that \(\hat{F}(p)\) is equal to the projection \(\hat{F}(R) \to F(R / \mathfrak {m}_R^n)\) composed with the natural isomorphism \(F \cong \hat{F}|_{\mathbf{C}_\Lambda }\). This is a simple check.

Apply lemma 7. Check that the property is stable by composition (note that the pullback is in \(Sets\)).

Apply the condition to \(p : A \to k\) (using that \(F(k)\) and \(G(k)\) are singletons).

By induction on the dimension of \(V\).

A surjection \(g\) is either an isomorphism or it can be factored as a composition of small extensions. The result is clear for isomorphisms and we can show that if \(g_1\) and \(g_2\) are two suitable morphisms such that \(p_{f, g_1}\) and \(p_{f, g_2}\) are surjective then \(p_{f, g_2 \circ g_1}\) is surjective as well.

In fact in this case \((H_1)\), \((H_2)\) and \((H_4)\) hold trivially since \(p_{f, g}\) is an isomorphism for any morphisms \(f\) and \(g\), because \(h_R\) is a hom-functor, and pullbacks in \(\mathbf{C}_\Lambda \) are pullbacks in \(\hat{\mathbf{C}}_\Lambda \). For \((H_3)\), it suffices to show the following linear isomorphism \(t_R^* \cong \mathfrak {m}_R / (\mathfrak {m}_R ^2 + \mathfrak {m}_\Lambda R)\).