I am looking for classical and elementary reason that why nonsingular cubics are not rational?
5 Answers
As there is no precision about the dimension or about the field, I will try to give a description in some cases. There are plenty of other nice results about cubics.
1) Dimension 1: A smooth nonsingular projective cubic curve is not rational. This is because the arithmetic genus, which is $g=(d1)(d2)/2=1$ is not equal to $0$. If you work over the complex field, you can see the points of your curve as a smooth surface definer over $\mathbb{R}$, and see that it is a torus, and not a sphere. You can also see it by computing the canonical divisor, which is trivial by adjunction.
2) Dimension 2: A smooth nonsingular projective surface over an algebraic closed is rational. First you have to show that the surfaces contains at least two disjoint lines $L_1$, $L_2$. Then, you construct a map from $L_1\times L_2$ to your surface: to any general pair of points $(p_1,p_2)$, with $p_i\in L_i$ you associate the unique point $q$ in your surface such that $p_1,p_2,q$ are collinear.
3) Dimension 2 over a nonalgebraically closed field: in general, the cubic is not rational. A simple reason could be that there is no point defined over your field. But even with a point, you can only get unirationality but not rationality. There are plenty of ways to study these cubics surfaces, for example with birational geometry (articles of Iskovkikh, Manin for example).
4) Dimension 3: a general smooth cubic threefold is not rational. This is much more complicate, but very interesting and comes from the intermediate Jacobian, see "The Intermediate Jacobian of the cubic threefold" by Clemens and Griffiths (Annals of Mathematics 1972).
5) Dimension >3: the question is open. There are plenty of rational smooth cubics but a general one should not be rational.

2$\begingroup$ You mean "a general cubic threefold" wrt Clemens and Griffiths? $\endgroup$ Apr 19, 2013 at 17:48

$\begingroup$ In case 3) (smooth cubic surfaces over a field which is not algebraically closed), irrationality can sometimes be proved by showing that the Chow group of $0$cycles of degree $0$ does not vanish. $\endgroup$ Apr 20, 2013 at 4:57

$\begingroup$ @Ashwath Arbindranath, general means that there is an open set in the moduli space of cubic threefolds where it is not rational. And you are right that the precise description of this set is given in Clemens and Griffiths. @ Chandan Singh: thanks for the other way of seeing the nonrationality. $\endgroup$ Apr 20, 2013 at 7:59

$\begingroup$ The funny thing is that the arithmetic genus of a singular (e.g. nodal) cubic curve is also 1, at least if it is assumed to be geometrically integral. And such curves are actually rational. But in the nonsingular case, the arithmetic and geometric genera agree, and the latter is a birational invariant (unlike the former). $\endgroup$– R.P.Apr 20, 2013 at 8:51

4$\begingroup$ @ Jérémy: I think Ashwath is pointing out a minor typo, and suggesting that you replace "surface" with "threefold" in 4). $\endgroup$ Apr 21, 2013 at 9:34
Jérémy already gave a complete answer but if you want a simple and direct way to see that a smooth plane cubic given by $f(x,y)=0$ is not rational, look at the differential form $dx/(\partial f/\partial y)$ and prove, using nonsingularity and the fact that the degree is three, that it is a nonzero differential form without zeros or poles. There is no such thing on a rational curve.

$\begingroup$ I'm not sure if this is more elementary, but if the field was rational, one would have polynomials $h(t)$ and $g(t)$ s.t. $f(h(t),g(t))$ = 0. $f(x,y) = y^2(x)(x1)(x\lambda)$ for an elliptic curve. It seems as if $\alpha$ is a root of $h(t)$ , that forces it to be a root of either $g$, $g1$, or $g\lambda$. But not a double root and hence a contradiction. I've not put in all the details, but I think that works and to my taste, it is more 'elementary' $\endgroup$– mehApr 20, 2013 at 16:14

$\begingroup$ @aginesky: You need to consider $g,h$ rational functions, not just polynomials. That's not to say there isn't an argument along those lines. $\endgroup$ Apr 20, 2013 at 17:01

$\begingroup$ @ voloch Good point. That is why I made it a comment and not an answer :) $\endgroup$– mehApr 21, 2013 at 14:33
here are my class notes for the special case of a fermat cubic. the same argument generalizes to all non singular cubics over algebraically closed fields of charcacteristic p = zero and maybe p > 3.
Assume that (x/z)^3 + (y/z)^3 = 1, where x, y,z are polynomial functions of t with no common factors. (since neither x/z nor y/z is constant, neither x nor y is zero.) Multiply through to obtain
(1): x^3 + y^3 = z^3, and differentiate to obtain
(2): x^2x' + y^2y' = z^2z', where ' denotes differentiation w.r.t. t.
Now we want to eliminate the z terms. multiply (1) by z' and (2) by z, and subtract, to obtain
z'x^3 + z'y^3 = x'x^2z + y'y^2z. Collecting terms, and factoring gives
x^2( xz'zx') = y^2 ( y'z  yz'). If either ( y'z  yz')=0, or ( xz'zx') = 0, then both do, and hence by the quotient rule for derivatives, we would have x/z and y/z constant.
Since x,y are rel prime, x^2 divides (yz'zy'), in ^[t], and thus 2.degree(x) ≤ deg(y)+deg(z)1. Repeating the argument for each of the other two variables, i.e. eliminating the x and y terms, leads to the same inequality with the variables permuted.
Adding the 3 inequalities gives 2.[deg(x)+deg(y)+deg(z)] ≤ 2.[deg(x)+deg(y)+deg(z)]  3, a contradiction. QED.
In the book 'Undergrad Algebraic Geometry' by Miles Reid, he proves the following theorem.
Let k be a field of characteristic not 2. Consider the curve $C$ defined (over $k$) by $C: y^2 = x(x1)(x \alpha)$ for $\alpha \in k  \{0,1\}$. Suppose you manage to find $f,g \in k(t)$ satisfying the equation of $C$. Then $f,g \in k$.
This proves you can't have a map $\mathbb{P}^1 \to C$, which proves that $C$ isn't rational.
Edit: I make the post because the proof is entirely elementary  it only uses basic properties of UFD's  so it is cool in that regard! Of course, this is only a special case of what is being asked for, since we are assuming the roots live in $k$, but it is still kind of neat.
If you work with an algebraically closed field of characteristic different than 2, a nonsingular cubic plane curve is isomorphic to an element of the Legendre family $y^2 z =x(xz)(x\lambda z)$ with $\lambda\neq 0, 1$.
The following automorphism of the projective plane $[x:y:z]\mapsto [x:y:z]$ induces an automorphism of $C_\lambda$ with 4 fixed points (these are the points of order 2 in the MordellWeil group of the elliptic curve).
This is enough to see that a nonsingular cubic curve cannot be rational as the automorphisms of $\mathbb{P}^1$ correspond to affine fractional linear transformation $t\mapsto \frac{at+b}{ct+d}$ and the only such transformation with more than 2 fixed points is the identity.