Affine Algebraic Curve cannot embed to Affine $n$-space

Solution 1:

The answer depends on what you take your definition of a curve to be and also what fields you work over.

If you assume that a curve is smooth and you're working over an infinite field, then every curve can be embedded in $\Bbb A^3$ for the same reasons every smooth projective curve can be embedded in $\Bbb P^3$: embed $X$ in some big $\Bbb A^n$, then project down, and as long as you project from a point not on the secant variety of $X$, you get an isomorphic curve. Since the secant variety is at most $2\dim X+1=3$ dimensional, you can always find a suitable projection.

If you allow singular curves, this result no longer holds. For instance, take an arbitrary curve and add some nilpotent elements at a closed point: we can make the tangent space to that point be of arbitrarily large dimension, and since closed immersions give injective maps on tangent spaces, the curve cannot embed in to any $\Bbb A^n$ with $n$ less than the dimension of the tangent space. Alternatively, we can take a smooth curve and glue together several points with distinct tangent directions to repeat the same sort of example.

If you're working over a finite field, you may also run in to cardinality issues. If $|X(\Bbb F_q)|>|\Bbb A^n(\Bbb F_q)|$, then $X$ cannot embed in $\Bbb A^n$ and indeed there are families of curves where the number of $\Bbb F_q$-rational points grows without bound (search "optimal tower" for concrete examples).