Show that a variety has degree 1 if and only if it is linear

I'm reading Karen Smith's Invitation to Algebraic Geometry and I'm stuck on the following question:

Show that a subvariety of $\Bbb{P}^n$ has degree one if and only if it is a linear subvariety.

The degree of an $m$-dimensional variety $V \subset \Bbb{P}^n$ is defined to be the maximum finite number of points of intersection of $V$ with a linear subvariety of codimension $m$ that doesn't contain $V$.

I can show that any linear subvariety has degree one. But, how do you show that a subvariety of degree one must be a linear variety? I don't know how to start, because I don't know any conditions that would be sufficient to show that the variety is linear.

• You want the linear subvariety to not contain $V$. Otherwise the number of intersection points is infinite! – user38268 Dec 12 '13 at 12:06

I'll treat the case $\dim V = 1$. To show a one-dimensional variety of degree $1$ is a line, we show that it is contained in any hyperplane passing through two points on $V$. So choose $x,y \in V$ with $x\neq y$ and let $H$ be a hyperplane containing $x$ and $y$. Notice $H$ is a linear variety of codimension $1$.
If $V \nsubseteq H$, the projective dimension theorem implies that every irreducible component of $V \cap H$ has dimension $0$. But now because $V$ has degree $1$, this means that $V \cap H$ can only contain $1$ point, contradicting $x,y \in V \cap H$. It follows $V$ is contained in every hyperplane passing through any two points of it, so that $V$ is a line (exercise ).