Question

Show that $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-2\right)$}\right.$is not isomorphic to$\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$. [ Hint: Exercises 2 and 5 may be helpful.]

Short answer

It is proved $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-2\right)$}\right.$is not isomorphic to $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$.

Step-by-step solution

Statement of Exercise 2 and Exercise 5

Exercise 2states that $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{\left.r+s\sqrt{2}\right|r,s\in \mathrm{\mathbb{Q}}\right\}$will be a subfieldof $\mathrm{ℝ}$,and $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$willbe an isomorphic to localid="1649238084879" $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-2\right)$}\right.$.

Exercise 5states $\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)=\left\{\left.r+s\sqrt{3}\right|r,s\in \mathrm{\mathbb{Q}}\right\}$willbe a subfieldof $\mathrm{ℝ}$, and $\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$will be isomorphic to localid="1649238099095" $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$.

Show that ℚxx2-2 is not isomorphic to ℚxx2-3

Theorem 5.11states that, for a field $F$and irreducible polynomial in $F\left[x\right]$, localid="1649237542460" $\raisebox{1ex}{$F\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(P\left(x\right)\right)$}\right.$ will be anextension field $K$of $F$, which contains a root oflocalid="1649237557388" $f\left(x\right)$.

According toexercise2 and exercise 5,localid="1649237601766" $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-2\right)$}\right.\cong \mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$and localid="1649237585518" $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.\cong \mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$,respectively. It is observed$x^2-3$contains a root in$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$according toTheorem5.11.

Assume that there is a root of $x^2-3$in $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)$: therefore, ${\left(a_1\sqrt{2}+a_0\right)}^2=3$. This implies $2a_1{a}_0\sqrt{2}+2a_1^2+a_0^2=3$. There is either$a_0=0$and$2a_1^2=3$or$a_1=0$, and $a_0^2=3$.

In either case, $\sqrt{3}\in \mathrm{\mathbb{Q}}$, which is a contradiction.

As a result, $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-2\right)$}\right.$does not have roots of the polynomial $x^2-3$; therefore, it is not isomorphic to $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$.

Hence, it is proved that$\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$is not isomorphic to $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.$.