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First of all I think the following statement is true: Given a polynomial $f\in \mathbb R[x,y]$ of degree $3$, there is always at least one solution $(a,b) \in \mathbb R^2$ for $f(a,b)=0$, right? What if I replace all $\mathbb R$ by $\mathbb Q$? Does the statement still remain true including the case at infinity? (Edit: The answer is no, see comments.)

Is there any criterion for $K$ such that there exists a polynomial in $K[x,y]$ of degree $3$ that does not have any roots in $K^2$ nor at infinity, i. e. $\mathbb P^2(K)$? Why are the cases $K=\mathbb C$ or $\mathbb R$ different from $\mathbb Q$?

Thrash
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  • For the case of coefficients in $\mathbb{R}$, we could let $x=y$ and thus get a cubic polynomial in one variable. It is well-known that every real cubic polynomial has a least one root (by a continuity argument). This argument breaks down if we restrict ourselves to rational roots, but you haven't quite said that when the polynomial has rational coefficients, you are only interested in excluding rational roots. – hardmath Aug 10 '17 at 21:49
  • Right, that was my thought, too (for $\mathbb R$). Right, by roots I mean respectively to the given field $K$, in this case $(a,b) \in \mathbb Q^2$. – Thrash Aug 10 '17 at 21:52
  • So for $\mathbb{Q}$ we should be satisfied by exhibiting a cubic polynomial in one variable with no rational roots? – hardmath Aug 10 '17 at 21:54
  • My bad, didn't read though carefully. (but yes, a polynomial in one variable is still a polynomial in 2 variables as well...) – platty Aug 10 '17 at 21:55
  • Sorry, my fault. I actually want to consider the projective closure, so $x^3-2$ does have the root $[0:1:0]$ at infinity. – Thrash Aug 10 '17 at 22:03
  • Selmer's cubic $3x^3+4y^3+5z^3$ has only $(0,0,0)$ for a rational solution. – sharding4 Aug 10 '17 at 22:26
  • Thank you, of course! But what if $K=\mathbb F_2$ or an arbitrary field? What conditions must the field fulfill to have roots for any polynomial over that field? – Thrash Aug 10 '17 at 22:31
  • @Thrash: Are you now asking about when a field is algebraically closed? – hardmath Aug 10 '17 at 23:02
  • No, $\mathbb R$ for example is not algebraically closed, but every real polynomial in two variables of degree $3$ has at least one root. Sorry, I meant for any polynomial of two variables of degree $3$. – Thrash Aug 10 '17 at 23:13
  • "For the case of coefficients in $\mathbb R$, we could let $x=y$ and thus get a cubic polynomial in one variable." By the way, this method doesn't work for $x^3-y^3+x^2+1$. – Thrash Aug 12 '17 at 13:15

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