Prove that there exists a subfield $\mathbb{F}$ of $\mathbb{R}$ such that $\mathbb{F}$ is isomorphic to $Q[x]/(2x^3 -3x^2 +6)$.
The given polynomial is Eisenstein at $3$ and hence irreducible over $\mathbb{Q}$. The quotient ring is therefore a field. The polynomial has two turning points at $0$ and $1$ and it’s positive at both turning points and hence it has only one real root and two complex roots. The quotient ring is not a sub field of $\mathbb{R}$. How then can we find a sub field of $\mathbb{R}$ isomorphic to the quotient ring? Please help.
Quotient Rings by ideals generated by irreducible polynomials
abstract-algebra
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Best Answer
$\Bbb Q[x]/(2x^3-3x^2+6)$ is an extension of $\Bbb Q$ obtained by adjoining a single root of $2x^3-3x^2+6$ to $\Bbb Q$. It is (a priori) not the splitting field of $2x^3-3x^2+6$.
Which is to say, $F$ doesn't have to contain all the roots, just one. Some times the other roots come along for the ride automatically, but in this case, at least if you pick the right root to start with, you can find a really good reason (pun intended) why this won't happen.