I have a question about exercise 2.8 from Hartshorne's "Algebraic Geometry" (see page 80):
Assume $X$ is a scheme over field $k$ and $x \in X$. Then the tangent space $T_x$ is defined as the set of $k(x) = O_{X,x}/m_x$-linear morphisms $m_x/m_x^2 \to k(x)$ (so the dual to $m_x/m_x^2 $).
We denote by $k[\epsilon]:=k[t]/t^2$ the ring of dual numbers.
I have to derive a 1-to-1 correspndence between elements of Tangent space $T_x$ and $k$-morphisms $f: Spec(k[\epsilon]) \to X$ with $f(\{*\})= x$ and $k(x)=k$ (so $x$ rational)
Here I encountered following PROBLEM:
At first glace it seems to be quite sraight forward:
A map $f: Spec(k[\epsilon]) \to X$ induces on stalks the map $f_x: O_{X,x} \to k[\epsilon]=k[t]/t^2$ and
since $f_x(m_x^2) =0$ (because $f_x$ is a local morphism since $f^{\#}$ is a morphism of local ringed spaces)
and the composition $m_x \subset O_{X,x} \to k[\epsilon]$ induces the map $m_x/m_x^2 \to k[\epsilon]$. If we futhermore quotient out $(t)$ from $k[\epsilon]$ then we obtain a canidate $m_x/m_x^2 \to k[\epsilon] \to k[\epsilon]/(t)=k$.
BUT: the problem is that this map is by construction a zero map since $f_*$ is again a local morphism so $f_x(m_x) \subset (t)$ and by $ k[\epsilon]/(t)=k$ we kill $(t)$.
So our $m_x/m_x^2 \to =k$ is always a zero map.
What is the error in my reasonings? Which construction is expected here to settle the desired correspondence?