Sorry to bother if this question is trivial: For a general smooth quintic threefold $V$ in $\mathbb{P}^4$ over an algebraic closed field $k$ of characteristic $0$. Is there a plane in $V$?
If Yes, how do I find it(them)? If no, under what condition there is a plane?
Thanks for any hints and comments.
There is also an elementary argument (I first saw it in http://www.math.stonybrook.edu/~jstarr/papers/appendix3.pdf):
Let $X=V(F)$ and suppose $X$ contains $P=\{X_0=X_1=0\}\subset\mathbb{P}^4$. Then, we can write $$F=F_0X_0+F_1X_1,$$ as $F\in \langle X_0,X_1\rangle$ by definition. Now, \begin{align*} \partial_0 F|_P &= F_0\\ \partial_1 F|_P &= F_1\\ \partial_2 F|_P &= 0\\ \partial_3 F|_P &= 0. \end{align*} Now, we notice that $F_0,F_1$ have a common zero when restricted to $P$, so $X=V(F)$ is not smooth.
In general, this shows we cannot have any planes in a hypersurface exceeding half the dimension of the hypersurface.