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Let $A$ be a local algebra over a field $k$. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. Suppose the canonical homomorphism $k \rightarrow A/\mathfrak{m}$ is an isomorphism. Let $f \in A$. There exists a unique $c \in k$ such that $f \equiv c$ (mod $\mathfrak{m}$). We denote this $c$ by $f(0)$. We call a $k$-linear map $v\colon A \rightarrow k$ a derivation if $v(fg) = v(f)g(0) + f(0)v(g)$ for $f, g \in A$. Let $Der(A, k)$ be the set of derivations. We regard $Der(A, k)$ as a vector space over $k$ in the obvious way. Let $v \in Der(A, k)$. If $f, g \in \mathfrak{m}$, $v(fg) = v(f)g(0) + f(0)v(g) = 0$. Hence $v(\mathfrak{m}^2) = 0$. Hence $v$ induces a $k$-linear map $\bar v\colon \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$. Hence we get a $k$-linear map $\psi\colon Der(A, k) \rightarrow (\mathfrak{m}/\mathfrak{m}^2)^*$. Is $\psi$ an isomorphism?

Motivation Let $k$ be the field of real numbers(or complex numbers). Let $M$ be a differentiable(or complex) manifold. Let $p \in M$. Let $\mathcal{O}_p$ be the ring of germs of differentiable(or holomorphic)functions at $p$. Then $\mathcal{O}_p$ is a $k$-local algebra satisfying the above condition. $Der(\mathcal{O}_p, k)$ is just the tangent space of $M$ at $p$.

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up vote 2 down vote accepted

Yes.

  1. Injectivity: let $v$ be a derivation such that $v(f)=0$ for all $f\in\mathfrak m$. As $v(k)=0$ (because $v(1)=v(1)1+1v(1)$ implies that $v(1)=0$), we have $v=0$.

  2. Surjectivity: let $\theta : \mathfrak m/\mathfrak m^2\to k$ be a linear form. Define $v: A\to k$ by $$v(f)=\theta(\overline{f-f(0)}).$$ Then $v$ is clearly $k$-linear. If $f, g\in A$, we write $f=c_1+f_1$, $g=c_2+g_1$ with $c_i\in k$ and $f_1, g_1\in \mathfrak m$. Then $fg=c_1c_2+c_1g_1+c_2f_1+f_1g_1$ and $$v(fg)=\theta(c_1\bar{g}_1+c_2\bar{f}_1)=c_1v(g)+c_2v(f). $$ This shows that $v$ is a $k$-derivation of $A$. Finally, by construction, $\psi(v)=\theta$.

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