Is it true that for every $n\in\mathbb{N}$ there exists an algebraic curve $C$ and a point $p$ in that curve whose tangent plane has dimension $n$? The title is very self-explanatory, I was thinking of finding a curve $k^n$ (k is the field with a non-singular point p, such that its tangent plane is V(0) and thus would be $k^{n}$ whose dimension is $n$.
 A: Consider the algebraic curve  embedded in affine $n$-space which is the union of all the co-ordinate axes. It is reducible, has a unique singular point at origin where the tangent space is $n$-dimensional.
A: P Vanchinathan's answer is absolutely correct (I just upvoted it) but his curve is reducible.
Here is an irreducible example:
Consider the curve $C\subset \mathbb A^n_k$ given parametrically by $(t^n,t^{n+1},\cdots,t^{2n-1})$.
It is irreducible since it is an image of $\mathbb A^1_k$.
However its tangent space at $O=(0,\cdots,0)$ is $T_O(C)=k^n$ : 
Indeed any polynomial polynomial vanishing in $C$, $f(x_1,\cdots,x_n)=q_1x_1+\cdots +q_nx_n+g(x_1,\cdots,x_n)\in I(C)$ (with $g$ containing only monomials of degree $\geq 2$) must satisfy $$q_1t^n+\cdots +q_nt^{2n-1}+g(t^n,\cdots,t^{2n-1})\equiv 0$$ for all $t\in k$
so that the linear form $q_1x_1+\cdots +q_nx_n$ must vanish, because all the exponents of $t$ appearing in $g(t^n,\cdots,t^{2n-1}$) are $\geq 2n$.
This proves that  $T_O(C)=k^n$, since  the tangent space is given by the vanishing of all the those linear forms obtained from the $f\in I(C)$.
