Is it true that every smooth variety (over $\mathbb{R}$ or $\mathbb{C}$ ) is a (real or complex) manifold? I have tried to show this using the implicit function theorem but I am not getting anywhere.
I think I understand it if I have a complete intersection variety,
If $X=V(f_{1},...,f_{m})\subset\mathbb{A}^{n+m}$ and $dimX=n$ then we have a function, $f:\mathbb{R}^{n+m}\to\mathbb{R}^{m}$, and the implicit function theorem gives a local function $g:\mathbb{R}^{n}\to\mathbb{R}^{m}$ such that $f(x,g(x))=0$. Thus, $X$ is the graph of $y=g(x)$.
The problem is that this uses the fact that $n+m$ is the dimension of the ambient space.
It suffices to work with affine varieties over a field $K$ (where $K={\mathbb R}$ or $K={\mathbb C}$). I will work with $K={\mathbb R}$ for concreteness. Let $V\subset {\mathbb R}^N$ be an affine subvariety. Smoothness of $V$ means that generators $f_1,...,f_n$ of its ideal $I_V$ satisfy the property that the rank $r$ of the derivative $DF$, $F=(f_1,...,f_n)$, is constant on $V$ (i.e. the Zariski tangent space of $V$ has constant dimension). Now, let us apply the Constant Rank Theorem, see say, here or here for a proof. The constant rank theorem states that for each point $x\in V$ there exists a smooth change of coordinates near $x$ and $F(x)$, after which the map $F$ locally has the form $$ F(x_1,...,x_N)= (x_1,...,x_r, 0,...,0). $$ Then, clearly, $F^{-1}(0)$ is a smooth submanifold in ${\mathbb R}^N$ of dimension $N-r$. This shows that $V$ is a smooth (real) manifold. In the case $K={\mathbb C}$ you use the holomorphic version of the constant rank theorem (proven in the same fashion as the real one; all what you need is the complex inverse function theorem) to conclude that $V$ is a complex submanifold.
Needless to say, complete intersections are irrelevant here.