Solutions in a field

Question

Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.

My attempt:

Suppose we have $x = b^{1/3} \in F$. We have that

$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = b^1 = b$$

Thus, there is one solution.

Assume there exists another solution and denote this by $c^{1/3}$. Since $b^{1/3}$ is a solution, $x = b^{1/3} \implies (b^{1/3})^3 = b \implies (b^{1/3})^3 - b = 0$. In a similar fashion, since $c^{1/3}$ is another solution, $x = c^{1/3} \implies (c^{1/3})^3 = b \implies (c^{1/3})^3 - b = 0$. Together we see that

$$(b^{1/3})^3 - b = 0 = (c^{1/3})^3 - b \implies (b^{1/3})^3 =(c^{1/3})^3 \implies b^{1/3} = c^{1/3} \implies b = c$$

Therefore there is at most one solution.

Is this correct for a field, $F$? Or did I make a mistake? Any assistance is much appreciated.

Answer 1

I think this only means that "one solution is unique", you still need to show $a_1^3>a_2^3$ if $a_1,a_2 \in F$

Answer 2

Just because you have $(b^{1/3})^3$ equals to $(c^{1/3})^3$, that doesn't mean that $(b^{1/3})$ equals $(c^{1/3})$. In fact, that's the very thing you're being asked to prove.

The theorem isn't true for all fields: for example, if $F=\mathbf{Z}_7$, then $x^3-1$ has three solutions, namely 1, 2 and 4.

However, you only have to prove it for ordered fields. $\mathbf{Z}_7$ is not an ordered field. Your proof will likely use the fact that the field is ordered.

Here's one such proof, but maybe you can find a simpler one:

Let $p$ and $q$ be two solutions to $x^3-b=0$, and let $p<q$ (note I'm using the fact that it's an ordered field) and let $q=p+d$.

Since $p^3-b=(p+d)^3-b=0$, it follows that $d(3p^2+3pd+d^2)=0$, but this equals $d((3p/2+d)^2+3p^2/4)$ (valid because $F$, being ordered, does not have characteristic $2$).

The only way a positive weighted sum of two squares can equal $0$ is if (because $F$ is ordered) both squares are $0$, so that means either $d=0$ or $3p/2+d=0$ and $p=0$, in which case $d=0$ anyway.

Thus $p^3=q^3$ if and only if $p=q$, hence $x^3-b=0$ has at most one solution for any $b$ (it may have no solutions).

There are probably simpler ways to prove this, but you will have to rely on the fact that the field $F$ is ordered. It's sufficient to prove that $f(x)=x^3$ is strictly monotonic.

Answer 3

You offer no justification for your claim that $b^{1/3} = c^{1/3} \implies b = c$.

In essence, that's what the problem requires you to prove.

As an indication that your proof is not valid, note that you never used the hypothesis that $F$ is an ordered field.

Thus, let $F$ be an ordered field, and suppose $x,y\in F$ are such that $x^3=y^3$.

We want to show $x=y$.

If $x=0$, then $x^3=0$, so $y^3=0$, hence $y=0$, so $x=y$.

Similarly, if $y=0$, then $y^3=0$, so $x^3=0$, hence $x=0$, so $x=y$.

Thus suppose $x,y\ne 0$, and assume $x\ne y$.

Our goal is to derive a contradiction.

\begin{align*} \text{Then}\;\;& x^3=y^3 \\[4pt] \implies\;& x^3-y^3=0 \\[4pt]\implies\;& (x-y)(x^2+xy+y^2)=0 \\[4pt] \implies\;& x^2+xy+y^2=0 \\[4pt] \implies\;& 4x^2+4xy+4y^2=0 \\[4pt] \implies\;& (4x^2+4xy+y^2)+3y^2=0 \\[4pt] \implies\;& (2x+y)^2+3y^2=0 \\[4pt] \end{align*} conradiction, since $(2x+y)^2 \ge 0$ and $3y^2 > 0$.