# Application of Field Extension to Linear Combination

## Problem 335

Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.

Let $\alpha$ be any real root of $f(x)$.

Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.

## Proof.

We first prove that the polynomial $f(x)=x^3-x+1$ is irreducible over $\Q$.

Since $f(x)$ is a monic cubic polynomial, the only possible roots are the divisors of the constant term $1$. As we have $f(1)=f(-1)=1\neq 0$, the polynomial has no rational roots. Hence $f(x)$ is irreducible over $\Q$.

Then $f(x)$ is the minimal polynomial of $\alpha$ over $\Q$, and hence the field extension $\Q(\alpha)$ over $\Q$ has degree $3$.

If $\sqrt{2}$ is a linear combination of $1, \alpha, \alpha^2$, then it follows that $\sqrt{2}\in \Q(\alpha)$. Then $\Q(\sqrt{2})$ is a subfield of $\Q(\alpha)$.

Then the degree of the field extension is

\begin{align*}

3=[\Q(\alpha): \Q]=[\Q(\alpha): \Q(\sqrt{2})] [\Q(\sqrt{2}): \Q].

\end{align*}

Since $[\Q(\sqrt{2}): \Q]=2$, this is impossible.

Thus, $\sqrt{2}$ is not a linear combination of $1, \alpha, \alpha^2$.

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