Let $\mathbb F$ be a field and let $a, b, c, d$ be fixed elements in the field $\mathbb F$. Consider the formulas

1) $\exists\;x\;\;:\;\;x^2=-1.$

2) $\exists\;x\;\;:\;\;(xa=c\land xb=d).$

Formula $(1)$ can be false in $\mathbb F$ but true in a field extension of $\mathbb F$. For exemple, formula $(1)$ is false in the reals $\mathbb R$ but true in the complex numbers $\mathbb C$.

The same does not happen with the formula $(2)$ since formula $(2)$ is true if and only if the matrix $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$ is singular, and it is well known that the singularity of a matrix is not changed by an extension of the field that contains the elements of the matrix.

My question is:

Is there any characterization of the formulas, in a field, whose validity does not change by an extension of the field?

  • $\begingroup$ What is $(a, b)$? $\endgroup$ – Asaf Karagila Nov 10 '13 at 15:06
  • $\begingroup$ $(a, b)$ can be interpreted as a vector in the space $\mathbb F^2$. $\endgroup$ – zacarias Nov 10 '13 at 15:58
  • 2
    $\begingroup$ It would be better to write it properly in the language of fields. That is $\exists x(xa=c\land xb=d)$. $\endgroup$ – Asaf Karagila Nov 10 '13 at 16:04
  • $\begingroup$ Not only "better", @AsafKaragila : it'd simply be clear! I didn't understand it as it was... $\endgroup$ – DonAntonio Nov 10 '13 at 16:56
  • $\begingroup$ Well, (1) involves a non-linear equation but (2) does not. This heuristic is not foolproof but is a good starting point. $\endgroup$ – Zhen Lin Nov 10 '13 at 18:01

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