= 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>, so
$$
\boxed{K_n=\mathbb Q[X]/(X^n-2)\text{ is a field}.}
$$
Let $\alpha$ denote the residue class of $X$. The <polynomial division algorithm> gives every residue class a representative of degree less than $n$, so $1,\alpha,\ldots,\alpha^{n-1}$ 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>, giving
$$
\boxed{\dim_{\mathbb Q}K_n=n.}
$$
The <quadratic polynomial> $X^2+1$ has no rational root, so it is irreducible over $\mathbb Q$. The same maximal-ideal argument proves \b[$L=\mathbb Q[X]/(X^2+1)$ is a <field>].
These two quadratic fields are \b[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 $\sqrt2$: evaluation at $\sqrt2$ 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 $K_2$.
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