A hyperplane is a hypersurface cut out by a single linear polynomial, thus a hypersurface of degree 1.

I have found the following mentionings of curves (in Hartshorne):

  1. An affine curve is defined to be $f(x,y)=0$ for $f$ an irreducible polynomial.

  2. A curve is a variety of dimension 1.

Using (1), a hyperplane is a special circumstance of a curve, since $f$ being linear implies that $f$ is irreducible.

Using (2) this implies that all hyperplanes are varieties of dimension $1$.

Moreover, this would imply that hyperplanes have degree and dimension equal to $1$.

However, in exercise 2.8 Hartshorne asks us to prove that a projective variety $Y \subset \mathbb{P}^n$ has dimension $n-1$ iff it is the zero set of a single irreducible homomogeneous polynomial of positive degree. We call such varieties hypersurfaces.

A hyperplane is a hypersurface and thus must have dimension $n-1$ by the above statement.

A hyperplane can also be considered a curve and thus must have dimension $1$.

These statements only work together when $n=2$.

Is it wrong to consider a hyperplane as a special type of curve in $\mathbb{P}^3$?

Something is going wrong in my implications since it only works out for $n=2$.

  • $\begingroup$ A hyperplane is always codimension 1 in the ambient space, whereas a curve is dimension 1, independently of the dimensionality of the ambient space. This, when the ambient space is dim 2, hyperplanes are curves. However if the ambient space is dimension 3, then a hyperplane would be dimension 2...etc $\endgroup$ – oxeimon Sep 6 '16 at 20:36
  • $\begingroup$ Your (1) describes an affine plane curve. He is not claiming to define the notion of a curve in general. A hypersurface has dimension one less than the ambient space, so is usually not a curve. It's not that they aren't related — in some sense they are dual — but they definitely are not the same $\endgroup$ – Hoot Sep 6 '16 at 20:36
  • $\begingroup$ Are curves (considered as projective varieties) then not the solution set of a single linear monomial then? $\endgroup$ – Jadwiga Sep 6 '16 at 20:38
  • $\begingroup$ Why monomial? Even without that: in higher dimensions this has no chance of being true. Krull's theorem tells you that a single equation cuts down the dimension by at most one. So there's no way to get from $\mathbb P^3$ to a curve inside of it by just imposing one condition. $\endgroup$ – Hoot Sep 6 '16 at 20:39
  • $\begingroup$ Ok. I understand what you are saying. The definition for an affine curve is not applicable here at all. Thank you. $\endgroup$ – Jadwiga Sep 6 '16 at 20:42

As Nesos' comment says, curves are any 1 dimensional objects, while hypersurfaces are codimension 1 objects - given by the vanishing of one equation. When ambient space is 2 dimensional (i.e. $\mathbb{A}^2$ or $\mathbb{P}^2$), hypersurfaces are curves, and indeed curves are hypersurfaces. But in higher dimensions, hypersurfaces are not curves.

Lubin's comment reminded me, hyperplanes are hypersurfaces given by the vanishing of a single linear polynomial (regardless of the dimension of the ambient space). So in $\mathbb{P}^3$, hyperplanes would be isomorphic to $\mathbb{P}^2$, and would not be curves.

  • $\begingroup$ And of course hyperplanes are flat. $\endgroup$ – Lubin Sep 6 '16 at 20:44

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