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Let $k$ be a field, $A$ be a Noetherian local $k$-algebra, $m$ its maximal ideal, and an isomorphism $i:A/m \to k$ . Let $v:m/m^2 \to k$ be a $k$-linear map (i.e. a Zariski tangent vector). I believe there exists a $k$-local homomorphism $f:A \to k[\epsilon]$ (where $\epsilon^2=0$) sending $x \in m$ to $v(x)\epsilon$. But how do I argue this rigorously?

Given $a \in A$, define $$f(a)=i(\overline{a}) + v(a-i(\overline{a}))\epsilon $$

How do I argue this is well-defined?

usr0192
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  • Why would your f be not well-defined? – Mariano Suárez-Álvarez Jun 23 '16 at 03:26
  • @MarianoSuárez-Alvarez Oh, you are right, it is well defined. In my mind I was thinking I was only giving it for generators of $m$, but that just not the case. Thank you! – usr0192 Jun 23 '16 at 03:38
  • @user26857 thanks for pointing out the error. It should be $v(x)\epsilon$. originally I had called the tangent vector $\alpha$ but then decided to change to $v$ since $\alpha$ looked too similar to $a$. – usr0192 Jun 23 '16 at 11:46

1 Answers1

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Consider the exact sequence $0\to m/m^2\to A/m^2\to A/m=k\to 0$. Use $\nu$ to get a push out $0\to k\epsilon\to k[\epsilon]\to k\to 0$. Check whatever is required for this map $A\to k[\epsilon]$.

Mohan
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