 Solution

The polynomial $X^n-2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with prime two: all coefficients except the leading one are divisible by two, while the constant coefficient is not divisible by four. Together with Gauss's lemma, this establishes irreducibility in $\mathbb{Q}[X]$. Part (b) shows that its principal ideal is a maximal ideal, soLet $\alpha$ denote the residue class of $X$. The polynomial division algorithm gives every residue class a representative of degree less than $n$, so $1,\alpha ,\dots ,{\alpha }^{n-1}$ span $K_n$ over $\mathbb{Q}$. If a linear combination of them vanished, its degree-less-than-$n$ representative would be a multiple of the degree-$n$ polynomial $X^n-2$, and hence would be zero. They are therefore a basis, givingThe quadratic polynomial $X^2+1$ has no rational root, so it is irreducible over $\mathbb{Q}$. The same maximal-ideal argument proves $L=\mathbb{Q}[X]/(X^2+1)$ is a field.

These two quadratic fields are not isomorphic. In $L$, the class of $X$ squares to $-1$. But $K_2$ embeds in $\mathbb{R}$ by sending its class of $X$ to $\sqrt{2}$: evaluation at $\sqrt{2}$ has kernel $(X^2-2)$, by irreducibility and polynomial division. No real number squares to $-1$, so no element of $K_2$ does. A field isomorphism preserves one, minus one and squaring, so it cannot carry $L$ to $K_2$.