Does $K = \mathbb Q[X]/(X^4 - 2)$ contain the imaginary unit $i$? 
Let $P(X) = X^4 - 2 \in \mathbb Q[X]$.
a) Prove that $P(X)$ is irreducible.
b) Prove that the field $K = \mathbb Q[X]/(P(X))$ is an algebraic extension of $\mathbb Q$ and find a generator of it.
c) Considering $K$ as a subfield of $\mathbb C$, determine if it contains the
complex number $i$.
d) Considering $K$ as a vector space over $\mathbb Q$, find a basis of $K$.


My solution
a) We have that $P(X) = (X + \sqrt[4]{2})(X - \sqrt[4]{2})(X^2 + \sqrt 2)$. But
$\sqrt 2 \notin \mathbb Q$ by the usual argument and the same can be said of
$\sqrt[4]{2}$. We conclude that $P(X)$ is irreducible over $\mathbb Q$.
My solution to points b),c),d) has been invalidated by a typo in the question.
Question
I think $a)$ is correct. Not sure about $b)$ and $d)$. Regarding $c)$, can someone give me a hint?
 A: For a); not having a root does not imply irreducibility. For example, $X^4+2X^2+1$ has no real roots, but it is certainly reducible. In stead consider using Eisenstein's criterion.
For b); it is not true that $\alpha=q+(P(X))$ for some $q\in\Bbb{Q}$. In stead, $\alpha=Q(X)+(P(X))$ for some $Q(X)\in\Bbb{Q}[X]$. If you have seen enough theory, you might say that because $K$ is a finite dimensional vector space over $\Bbb{Q}$, the consecutive powers $1,\alpha,\alpha^2,\ldots$ cannot be linearly independent. A nontrivial relation between them yields a polynomial of which $\alpha$ is a root.
For c); there is more than one way to consider $K$ as a subfield of $\Bbb{C}$. One is a real embedding given by mapping $X$ to $\sqrt[4]{2}$. Then $K$ does not contain $i$ because $\Bbb{R}$ does not contain $i$.
For d); the argument for b) show that your argument here fails. Give it some more thought after having understood the previous questions, I think you can figure this one out now.
A: Your argument of a) is incomplete. You proved that $P(X)$ has no root in $\mathbb Q$, and to draw the conclusion it's necessary to add the proof that $P(X)$ has no factor of degree $2$, which is pretty similar to your original argument. In fact, if you know that a monic polynomial $f(X)\in\mathbb Z[X]$ is irreducible iff it's irreducible in $\mathbb Q[X]$, it'll be more plain to check that the image of  $P(X)$ in $\mathbb Z_4[X]$ is irreducible.
For d), seems that you were messing the notion of a generating set of a  ring and that of a  $\mathbb Q$ vector space. In fact it's easy to see we can write any element in $K$ as a $\mathbb Q$ linear combination of $1,\bar X,\bar X^2,\bar X^3$ ($\bar X$ is the image of $X$ in $K$) and the four elements are $\mathbb Q$ linearly independent,thus form a basis and the $\mathbb Q$ dimension of $K$ is $4$.
For b), call a field extension $E/F$ finite if $E$ is a finite dimensional $F$ vector space (in case you don't know, $E$ is always a vector space over $F$, so the main concern lies in the dimension, not being a vector space). It's not hard to prove that finite field extensions are always algebraic.
For c), define a morphism $\tilde i:\mathbb Q[X]\to\mathbb R, X\mapsto\sqrt[4] 2$, it's routine to check that $\ker\tilde i\supseteq(P)$, thus $\tilde i$ induces $i: K\to\mathbb R$. But a ring morphism between two fields must be injective, thus we can regard $K$ as a subring of $\mathbb R$. Since $-1$ has no square root in $\mathbb R$, neither does it in the subring $K$.
A: I suppose that there are two misunderstandings. The first is, that you showed  $[K:\mathbb{Q}]=1$. This would imply that $K$ is isomorphic to $\mathbb{Q}$, and hence does not contain $i$. More probably, however, is that the field $K$ is not $K = \mathbb Q/(X^4 - 2)$, but rather $K = \mathbb Q[x]/(X^4 - 2)$. Then we can work with the quotient. Because of Eisenstein, $x^4-2$ is irreducible, so that the ideal $(x^4-2)$ is maximal, and we have that the quotient is a field. Clearly we have $[K:\mathbb{Q}]=4$ then, with $K=\mathbb{Q}(2^{1/4})$. This is a real field, not containing $i$. It has the basis $1,a,a^2,a^3$ with $a=2^{1/4}$.
