# $\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than 3 with $P(\sqrt[3]{2})=0$ [duplicate]

$$\sqrt[3]{2}$$ satisfies $$x^3-2=0$$ Show that there is no polynomial $$P(x)$$ of degree less than $$3$$ with $$P(\sqrt[3]{2})=0$$ All coefficients are rational numbers.

Is it by induction? Say, if $$x$$ has degree of $$1$$, it doesn't work; and for $$x$$ having degree of $$2$$, am I applying Polynomial remainder theorem?

• Welcome to Maths SX! Do you mean $3\sqrt 2$ or $\sqrt[3]2$? Oct 7 '19 at 19:24
• You mean $\sqrt[3]{2}$. And this is the minimal polynomial, see the duplicates. Oct 7 '19 at 19:25
• As stated, it’s false whether you meant $3\sqrt2$ or $\sqrt[3]2$, since whatever value $a$ you have, it’s a root of $P(x)=x-a$. Perhaps there’s some restriction on the coefficients of the polynomials that you haven’t mentioned?
– amd
Oct 7 '19 at 19:28
• You need to specify the field your polynomial is over. For example the polynomial $x-2^{1/3}$ would be a counterexample to your claim if the field was arbitrary. I assume the field is $\Bbb{Q}$. As for your task, just factor $(x-2^{1/3})$ out from your polynomial and show that the some coefficient of the other factor is not in $\Bbb{Q}$. Oct 7 '19 at 19:32
• Look up minimal polynomial. The polynomial $x^3-2$ is irreducible over $\Bbb{Q}$, so it must be the minimal polynomial of $\root3\of2$, i.e. the lowest degree polynomial with rational coefficients vanishing at $\root3\of2$. By the way, you need to specify the field of coefficients. See amd's comment as well as that of B.Swan. Oct 7 '19 at 19:33

You didn't precised what are the nature of the coefficients of P, so we can build a polynomial $$P(x) = x - 3\sqrt{2}$$ such that $$P(3\sqrt{2}) = 0$$.

And, we can build an infinity of polynomials of degree 2 such that $$P(2 \sqrt{3}) = 0$$, these polynomials are : $$P(x) = (x - 2 \sqrt{3})(x - k), k \in \mathbb{C}$$

We all must be very careful to mathematical rigor.

If we consider that $$P \in \mathbb{Z}[X]$$, it is obvious that there is no polynomials of degree 1 such that $$P(2 \sqrt{3}) = 0$$ because $$\forall P \in \mathbb{Q}_1[X], P(X) = 0 \implies X \in \mathbb{Q}$$, and, because $$\mathbb{Z}_1[X] \subset \mathbb{Q}_1[X]$$, this property is true even for polynomials with relative coefficients (and $$2 \sqrt{3} \notin \mathbb{Q}$$).

Then, for polynomials with degree 2, we use the quadratic formula :

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, so we search solution to the equation :

$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = 2 \sqrt{3}, (a,b,c) \in \mathbb{Z}^3$$

We can directly see that we must have $$b = 0$$ because if that was not the case, we should have $$x = k + l \sqrt{m}, k \neq 0$$ knowing that b must be a relative number. Or, we see that the solution we are searching for is not looking like that.

So, because we have $$b = 0$$, we have $$x = \frac{\pm \sqrt{-4ac}}{2a}$$ with one of these two solutions equal to $$2 \sqrt{3}$$. Because we have $$2\sqrt{3} \notin \mathbb{C} - \mathbb{R}$$, we must have $$-4ac > 0$$ so $$4ac < 0$$.

We have so $$x = \pm 2 \frac{\sqrt{-ac}}{2a} = \pm \frac{\sqrt{-ac}}{a}$$ Then, we suppose $$a > 0$$ so we can write $$x = \pm \frac{ \sqrt{a} }{a} \times \sqrt{-c} = \pm \frac{\sqrt{-c}}{\sqrt{a}}$$ so we must have $$\sqrt{-c} = \sqrt{3}$$ and $$1/\sqrt{a} = \pm 2$$ which is not possible with $$a \in \mathbb{Z}$$

And, if we have $$a < 0$$, we have $$x = \pm \frac{\sqrt{c}}{\sqrt{-a}}$$

So we must have $$c = 3$$ and $$1/\sqrt{-a} = \pm 2$$ which is not possible with $$a \in \mathbb{Z}$$, QED.

EDIT : The proof is also valable for $$x = 2^{1/3}$$ and $$3\sqrt{2}$$