Is there an easy example of a variety that is not rational? ${}$

Question

I saw a lot of examples of varieties which are rational, now I was also wondering if there was a simple example of a variety that is not rational. Sadly I didn't find that much online, is it maybe hard to find them?

Answer 1

HINT:

You can check that in char $\ne 3$ the curve $x^3 + y^3 = 1$ is not rational.

$\bf{Added:}$

By a sign change, reduce to $x^3 + y^3 + 1 = 0$ (easier to deal with). Now, assume that we have a rational parametrization $(x,y) = (\frac{P(t)}{R(t)}, \frac{Q(t)}{R(t)})$, with $P$, $Q$, $R$ polynomials. We get $$P^3(t) + Q^3(t) + R^3(t) = 0$$

Assume that $\operatorname{gcd}(P, Q, R) = 1$, otherwise divide by the common divisor. Now, since the parametrization is non-trivial, at least one of the $P$, $Q$, $R$ is non-constant. Take the derivative in the equality above and get $$3 P'(t) P^2(t) + 3 Q'(t) Q^2(t) + 3 R'(t) R^2(t) = 0$$

Since the char is $\ne 3$, we may divide by $3$. Now we get two linear equations in the vector $(P^2(t), Q^2(t), R^2(t))$. It follows that this vector is proportional to $$(P(t), Q(t), R(t)) \times (P'(t), Q'(t), R'(t))$$ Note that this above vector is $\ne 0$, otherwise $P$, $Q$, $R$ have constant quotients, not possible. Also note that the vector $(P^2(t), Q^2(t), R^2(t))$ has relatively prime entries. It follows that $$M(t) \cdot (P^2(t), Q^2(t), R^2(t) ) = (P(t), Q(t), R(t)) \times (P'(t), Q'(t), R'(t))$$ where $M(t)$ is a polynomial. But now we have problems with the degree, since if $P(t)$ is then the polynomial of largest degree then $M(t) \cdot P^2(t) \not = Q(t) R'(t) - R(t) Q'(t)$, contradiction.

Note: This works for the curve $a x^n + b y^n +c = 0$, $a b c \ne 0$, $n\ge 3$, char $\not \mid n$.

$\bf{Added:}$ Let $A$ be a UFD ( in our case $k[t]$, where $k$ is the basic field), $K$ its field of fractions. Let $m\ge 2$ be a natural number, and consider $K^m$, $m$-size vectors with components in $K$ ( here $m=3$). Now, let $v \in A^m\subset K^m$, with the GCD of its entries $1$, and let $A^m \ni w = \alpha v$ proportional to $v$. Then $\alpha \in A$. (for us $v = (P^2, Q^2, R^2)$, $w = (P,Q,R) \times (P', Q', R')$ ).

Indeed, we have $w_i = \alpha v_i$ for all $i$. If $e_p(\alpha) < 0$ for some prime of $A$, and $i$ is one of the components for which $e_p(v_i) =0$ ( the components of $v$ are relatively prime ), we would get $e_p(w_i) < 0$, contradiction. Another proof (if the domain is PID, like $k[t]$) take $\sum c_i v_i = 1$, and so $\alpha = \alpha \cdot 1 = \alpha \cdot (\sum c_i v_i) = \sum c_i w_i \in A$.