How to write down a non-degenerate cubic surface in P^4 I want to do something very concrete: write down a smooth scheme of given degree and dimension in projective n-space. 
A natural way to go about this is to try and write down a complete intersection, but not all degrees/dimensions can be gotten this way. For instance, I want to write down a smooth, non-degenerate cubic surface in $\mathbb{P}^4$.
What's a systematic way to go about this kind of problem?
 A: I don't know what a systematic method would be, but in this particular case, the Segre embedding $\mathbb{P}^2 \times \mathbb{P}^1 \to \mathbb{P}^5$ has degree $3$. So, by Bertini's theorem, a generic hyperplane slice of $\mathbb{P}^2 \times \mathbb{P}^1$ will be a degree $3$ surface in $\mathbb{P}^4$.
A: Let me try to generalise David Speyer's construction a little to show how one can achieve a certain range of degrees and dimensions.
First let me change the notation a little bit: rather than asking for a subvariety $X$ of dimension $b$ and degree $d$ in $\mathbf P^n$, let's ask for a projective vareity $X$ of dimension $b$, codimension $c$, and degree $d$. (Of course the data $(b,n,d)$ and $(b,c,d)$ are equivalent information; I just find it easier to write down the answer in terms of the latter.) Let's call such a variety $X_{b,c,d}$.
Then my claim is the following:
Claim: For any $d \geq 2$, there exist smooth projective varieties $X_{b,c,d}$ in the following cases:


*

*$b=d$ and $c=d-1$;

*$b=d-1$ and $c=d-1$;

*$b<d-1$ and $b+1 \leq c \leq d-1$.


Remark 1: The maximum codimension of a nondegenerate subvariety of degree $d$ in projective space is $d-1$. This explains the bound on $c$ above.
Remark 2: The claim says we can always get what we want as long as the codimension $c$ is (roughly) at least as big as the dimension $b$. What about when codimension is small compared to dimension? Well, Hartshorne's conjecture predicts that when $b > \frac{2}{3}(b+c)$, in other words when $c<\frac12 b$, any smooth $X_{b,c,d}$ must be a complete intersection. So one should not expect to be able to do this for all $c$. On the other hand, there is plenty of room between $\frac12 b$ and $b+1$; maybe one could do something clever in that range, but I have no idea. It seems much harder.
Proof of Claim: Suppose $b,c,d$ are given. To start with, let $S_d$ be the embedding of $\mathbf P^1 \times \mathbf P^{d-1}$ in $\mathbf P^{2d-1}$ given by the line bundle $O(1,1)$. One can check this has degree $d$, so it is an $X_{d,d-1,d}$ as in the first bullet point. As in David Speyer's answer, one can then take a general linear section of this to get an $X_{d-1,d-1,d}$ as in the second bullet point. 
Now assume $b<d-1$ and $b+1 \leq c \leq d-1$. Start with the $X_{d-1,d-1,d}$ as above. Take a section by a general linear subspace of the right dimension to get $X_{b,d-1,d}$. This has the right degree and dimension, but the wrong codimension. To fix this, we project away from a point not on $X$, repeatedly; this lowers the codimension by 1 at each stage. As long as $c>b+1$, the secant variety of $X_{b,c,d}$ is not all of $\mathbf P^{b+c}$, so we can project like this and get a smooth $X_{b,c-1,d}$. 
