Question

prove that Q[x]/<x-2> is isomorphic to Q as fields.

Answer 1

Polynomial remainder theorem implies that each element of $\mathbb Q[x]/\langle x-2\rangle$ is of the form $[f(x)]=f(x)+\langle x-2\rangle=f(x)+(x-2)\mathbb Q[x]$, where $\deg f(x)<\deg(x-2)=1$, and therefore, $f(x)\in\mathbb Q$. Consequently, $\mathbb Q[x]/\langle x-2\rangle=\{[a]\ : \ a\in\mathbb Q\}$.

Consider the map $\psi: \mathbb Q\to \mathbb Q[x]/\langle x-2\rangle, \psi(a)=[a].$ Taking into account that $\psi(ab)=[ab]=[a][b]=\psi(a)\psi(b)$ and $\psi(a+b)=[a+b]=[a]+[b]=\psi(a)+\psi(b)$, we conclude that $\psi$ is a field homomorphism. If $a\ne b$, then $\psi(a)=[a]\ne [b]=\psi(b)$, and consequently $\psi$ is injective. Since $\psi(a)=[a]$ for any $[a]\in \mathbb Q[x]/\langle x-2\rangle$, $\psi$ is surjective. Therefore, $\psi$ is a field isomorphism. Thus the fields $\mathbb Q$ and $\mathbb Q[x]/\langle x-2\rangle$ are isomorphic.