It's an exercise in the Wedhorn's book:
Let $k$ be an algebraically closed field. Give a $k$-scheme $X$ such that:
There is a morphism $f:\mathbb A_k^1\rightarrow X$ which is homeomorphic on the topological space.
$\dim T_xX=1$ at all except one $x\in X$.
$X$ is not reduced.
I have tried $k[X,Y,Z]/(X^2+Y^3,Z^2)$. In this case, we can calculate its jacobbian, and see that the tangent space has dimension $3$ at $0$ but $2$ elsewhere. But in the case $k[X,Y]/(X^2-Y^3)$ it satisfies all but condition 3.
My idea is: Since we need to get a "special" point, we may construct a curve which is singular at only one point. (In above cases $X^2-Y^3$ is the curve I need). But I don't know how to modify it to make it satisfy all conditions. Could you help me achieve it? Or could you give an example directly? Thanks!
Hopefully this will work.
Consider $k[x,y]/(xy,y^2)$.
The non-reduced point is the origin. It is geometrically the $x$-axis with only the origin thickened.
There is a reduction morphism $k[x,y]/(xy,y^2)\to k[x]$ given by killing the nilpotent ideal $(y)$. This induces a morphism $\mathbf{A}^1_k\to \operatorname{Spec}k[x,y]/(xy,y^2)$ which must be a topological isomorphism.
The Jacobian, $J(xy,y^2)$, is given as
$$J(xy,y^2)=\begin{pmatrix} y & x \\ 0 & 2y \end{pmatrix}:k^2\to k^2.$$
All closed points lie along the $x$-axis so that at all points other than the origin the rank of the matrix is $1$ and at the origin the rank is $0$.
Note that there are no characteristic requirements on $k$.