Let $F$ be a certain field.

Prove or disprove that the following statements:

  • The equation $X^3=0_F$ has only one solution.

  • The equation $X^3=1_F$ has only one solution.

  • Suppose F is finite, then the equation $X^3=1_F$ has only one solution.

I'm pretty sure all of them are true but I'm not sure how to write the proof so any help would be appreciated.

Note: I'm only in my first few weeks of linear algebra so I don't think any advanced solutions would help.


$X^3 = 0$ has only one solution. This is because fields do not have zero divisors, which follows from requiring every element to be invertible.

$X^3 = 1$ might have more solutions. In $\mathbb{C}$, for example, there are three cube roots of unity.

For finite fields we can, again, have more than two solutions. Consider $\mathbb{F}_7$ - here both 1 and 2 are solutions, because $2^3 = 8 = 1 \;mod\; 7$.

  • $\begingroup$ Thanks! BTW is there a particular reason why you chose mod 7 ? I know that basically any mod can work here but I'm just curios if there's something special with mod 7. $\endgroup$ – GinKin Nov 20 '13 at 18:55
  • 1
    $\begingroup$ Well it wouldn't be a field if it wasn't prime. Obviously 1 cubed is one, so I looked at 2 cubed for another solution to X^3 = 1. 2^3 not equal to 1 mod 2, mod 3 or mod 5, so in some sense 7 is the simplest example, but it isn't special. $\endgroup$ – Harry Wilson Nov 21 '13 at 15:21
  • $\begingroup$ It wouldn't be a field if it wasn't a prime ? Why is that ? $\endgroup$ – GinKin Nov 21 '13 at 20:41

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