could you please confirm my method of solution for the following question? Thanks in advance!! :)
Consider the polynomial $f(x) = x^3 + x^2 - 2x + 1 \in \mathbb Z[X]$. Prove that $f $ is irreducible in $\mathbb Q[X]$.
Solution: I tried to use Einstein's Criterion but it clearly does not work here. So I tried to solve for $x$ by using the Tschirnaus Transformation and turn it into a compress cubic and solve for $x$. I got $x_1, x_2, x_3 = $ some number that consists of nested square roots and the complex number i.
In this case, isn't it sufficient to show that $f$ is irreducible since the solution $x_1, x_2, x_3$ are clearly not in the set of Rationals?
Thanks! :)
If $\frac{p}{q}$ is a rational solution of $x^3+x^2-2x+1$ with $p,q\in \mathbb{Z}$ and $p,q$ coprime then
$$\frac{p^3}{q^3}+\frac{p^2}{q^2}-2\frac{p}{q}+1=0 \Leftrightarrow p^3+p^2q-2pq^2+q^3 = 0\Leftrightarrow p(p^2+pq-2q^2) = -q^3.$$ This implies that $p|q$. But since $p$ and $q$ are coprime it follows $p=\pm 1$. In the same way
$$\frac{p^3}{q^3}+\frac{p^2}{q^2}-2\frac{p}{q}+1=0 \Leftrightarrow p^3+p^2q-2pq^2+q^3 = 0\Leftrightarrow -p^3 = q(p^2-2pq+q^2).$$ This implies $q|p$. But since since $p$ and $q$ are coprime it follows $q=\pm 1$.
Hence, if there is a rational root of $x^3+x^2-2x+1$ it must be $\pm 1$ which is obviously not a solution. Consequently there is no rational number $\frac{p}{q}$ such that $(x-\frac{p}{q})\mid ( x^3+x^2-2x+1)$ which proves that $x^3+x^2-2x+1$ is irreducible.