Answer to Question #151281 in Abstract Algebra for Sourav Mondal

Question

Q[x]/<x-5> isomorphic to Q as field.

Answer 1

The elements of the field $\mathbb Q[x]/\langle x-5\rangle$ have the form $[f(x)]=f(x)+(x-5)\mathbb Q[x]$. According to the Polynomial Euclidean Algorithm for any polynomial $f(x)\in\mathbb Q[x]$ there is a unique polinomial $r(x)\in \mathbb Q[x]$ such that $f(x)=(x-5)q(x)+r(x),\ \deg(r(x))<\deg(x-5)=1,$ for some unidue polinomial $q(x)\in\mathbb Q[x]$. It follows from $\deg(r(x))<1$ that $r(x)=r\in\mathbb Q$, and thus $[f(x)]=[r].$ Therefore, $\mathbb Q[x]/\langle x-5\rangle=\{[r]\ :\ r\in\mathbb Q\}$.

Let us define a map $f:\mathbb Q\to\mathbb Q[x]/\langle x-5\rangle, \ \ f(r)=[r].$ Taking into account that $f(a+b)=[a+b]=[a]+[b]=f(a)+f(b)$ and $f(a\cdot b)=[a\cdot b]=[a]\cdot [b]=f(a)\cdot f(b)$, we conclude that $f$ is a field homomorphism. Since the reminder is a unique, $a\ne b$ implies $[a]\ne [b]$, and therefore, $f$ is injective. For each $[r]\in\mathbb Q[x]/\langle x-5\rangle$ we have that $f(r)=[r]$, and $f$ is surjective. Consequently, $f$ is a field isomorphism, and $\mathbb Q[x]/\langle x-5\rangle$ isomorphic to $\mathbb Q$ as field.