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I was reading about first order deformations from Hartshorne's Deformation Theory some time ago, and was thrown off by a small mistake in one of the exercises. Exercise 4.1 in section 4 of chapter 1 asks to prove that if a $k$-algebra endomorphism $f\colon A\to A$ of a local $k$-algebra with residue field $k$ induces an isomorphism $\bar{f}\colon A/\mathfrak{m}^2\to A/\mathfrak{m}^2$, then $f$ itself is an isomorphism.

Proof sketch

Firstly, since $f$ is a $k$-algebra morphism, we obtain a diagram with exact rows. This tells us that it suffices to show the restriction $$f\vert_{\mathfrak{m}}\colon\mathfrak{m}\to\mathfrak{m}$$ is an isomorphism. Now, taking quotients by the ideal $\mathfrak{m}^2$ in the above diagram gives Since $\bar{f}$ is an isomorphism, so is the restriction $\mathfrak{m}/\mathfrak{m}^2\to\mathfrak{m}/\mathfrak{m}^2$. The obvious thing to do here is to try applying Nakayama's lemma.

Proving the surjectivity of $f\vert_{\mathfrak{m}}$ is a breeze. The surjectivity of $\bar{f}\vert_{\mathfrak{m}/\mathfrak{m}^2}\colon\mathfrak{m}/\mathfrak{m}^2\to\mathfrak{m}/\mathfrak{m}^2$ implies $f(\mathfrak{m})+\mathfrak{m}^2=\mathfrak{m}$, which implies by Nakayama's lemma that $f(\mathfrak{m})=\mathfrak{m}$. Injectivity does not follow quite as easily. First, I claim that for $x\in\ker f\vert_\mathfrak{m}$, $x\in\mathfrak{m}^i$ implies $x\in\mathfrak{m}^{i+1}$. Indeed, for such an $i$ there are $x_1,\ldots,x_i\in\mathfrak{m}^i$ such that $x=x_1\cdots x_i$. Then, $$0=f(x)=f(x_1)\cdots f(x_i).$$ Now, we must have $f(x_j)\in\mathfrak{m}^2$ for some $j$, because otherwise $f(x_1)\ldots f(x_i)\in\mathfrak{m}^i\setminus\mathfrak{m}^{i+1}$ but this set does not contain $0$. The injectivity of $\bar{f}$ implies that $x_j\in\mathfrak{m}^2$, so $x\in\mathfrak{m}^{i+1}$. Using this claim inductively gives us that $$\ker f\vert_\mathfrak{m}\subseteq\bigcap_{i>0}\mathfrak{m}^i.$$ Finally, Krull's Intersection theorem implies that this intersection is the zero ideal, which completes the proof. As a remark, Krull's Intersection theorem uses Nakayama's lemma, so we relied on it in the proof of injectivity also.

Finiteness assumptions

The keen-eyed among you spotted that I didn't address some finiteness assumptions I should have. In particular, applying Nakayama's lemma in the proof of surjectivity assumes $\mathfrak{m}$ is finitely generated. Krull's Intersection theorem also requires that $A$ is Noetherian. The part where I got stuck in this problem was trying to give a general proof that works for non-Noetherian rings as well. After some time I realised that there is an easy non-Noetherian counter-example to the statement.

Counter-example

For a counter-example, we take a local $k$-algebra $(A,\mathfrak{m})$ which has residue field $k$ but is not itself $k$ and for which $\mathfrak{m}^2=\mathfrak{m}$. Given such a ring, we can define a $k$-algebra homomorphism $f\colon A\to A$ by fixing $k$ and collapsing $\mathfrak{m}$ to $0$. Since $\mathfrak{m}^2=\mathfrak{m}$, we have $A/\mathfrak{m}^2\cong k$, so $\bar{f}\colon A/\mathfrak{m}^2\to A/\mathfrak{m}^2$ is the identity map on $k$, while $f$ is not an isomorphism.

It is not difficult to construct such a ring. Consider for example the quotient $$k[x_0,x_1,x_2,\ldots]/I$$ of the polynomial ring in infinitely many variables by the ideal $$I:=(x_i-x_{i+1}^2\mid i\in\mathbb{N}).$$ The maximal ideal $$\mathfrak{m}=(x_0,x_1,x_2,\ldots)$$ clearly satisfies $\mathfrak{m}^2=\mathfrak{m}$. The ring is not local, but we simply take the localisation and set $$A:=\left(\frac{k[x_0,x_1,\ldots]}{I}\right)_\mathfrak{m}.$$ It is easy to see that $A/\mathfrak{m}\cong k$. It remains to show $A\ncong k$.

Assume for a contradiction that $A\cong k$. Then, say $x_0=\lambda$ in $A$ for some scalar $\lambda\in k$. This holds if and only if there is $s\in S:= k[x_0,x_1,\ldots]\setminus\mathfrak{m}$ such that $$s(x_0-\lambda)\equiv0\pmod{I}.$$ Since the polynomials in $S$ are the ones with non-zero constant coefficients, while the polynomials in $I$ have constant coefficient zero, we must have $\lambda=0$.

Now, suppose $x_0s\in I$, so there are $i_1,\ldots,i_r\in\mathbb{N}$ and $g_\ell\in k[x_0,x_1,\ldots]$ such that $$x_0s=\sum_{\ell=1}^r g_\ell(x_{i_\ell}-x_{i_\ell+1}^2).$$ Since $s$ has a non-zero constant term, $0$ must be among the indices $i_1,\ldots,i_r$. Assume without loss of generality that $i_1=0$. Then, $$x_0(s-g_0)=-g_0x_1^2+\sum_{\ell=2}^rg_\ell(x_{i_\ell}-x_{i_\ell+1}^2).$$ Since $x_0$ divides the left hand side, it must divide each $g_\ell$ as $x_0$ does not divide $x_{i_\ell}$ or $x_{i_\ell+1}$ for $\ell>1$. It follows that $$s=\sum_{\ell=1}^r\frac{g_\ell}{x_0}(x_{i_\ell}-x_{i_\ell+1}^2),$$ which is a contradiction since $s$ has a non-zero constant term, while the right hand side does not have a constant term. Thus, we arrive at a contradiction, so $sx_0\not\in I$ and $A\ncong k$.