Philosophy of simple field extensions

In B. L. van der Waerden's Algebra stuck on the problem 6.9:

The polynomial $f(x) = x^4 + 1$ is irreducible in the field of rationals. Adjoin a root $\theta$ and resolve the polynomial in the extended field $\mathbb Q (\theta)$ into prime factors.

Seems like I haven't nailed the idea of extending fields, so I'm asking to control my thoughts and help with troubled places.

We can obtain the desired extension by two ways: either we use the fact, which states that we know, in which field would that polynomial have a root (nonsymbolic adjunction), or we can build residue class field modulo that polynomial (symbolic adjunction).

I don't realize what I'm supposed to do in both cases.

1. In nonsymbolic way, should I just completely factorize $f(x)$ over $\mathbb C$, or am I to find only one (any shall do) root?
   I can write
   $f(x) = (x^2-i)(x^2+i) = (x-\sqrt i)(x + \sqrt i)(x - \sqrt{-i})(x + \sqrt{-i})$
   but what would that give me?
   
   Or I could simply say that $\theta = \sqrt{\sqrt{-1}}$ is an obvious root, divide $f(x)$ by $(x - \theta)$ and think hard what to do next with the quotient $x^3 + \theta x^2 + {\theta}^2 x + {\theta}^3$?

2. In symbolic way, I have to find residue class field modulo $f(x)$. Is this possible at all? I haven't seen yet a single example of making that with polynomials.

I'm really sorry for the size of the question. The systematic ignorance reminds of itself. However, I feel, that a proper answer will amend many other problems in my knowledge.

I am really grateful at least for the time you spent on reading it.

I have to admit that I have hard time to really understand the philosophy behind your question. Let's try however to provide feedback.

For whatever field $K$ and an irreducible polynomial $f \in K(x)$, $(f) \subseteq K(x)$ is a prime ideal and $K(x)/(f)$ a field. Moreover if $\overline K$ is an algebraic closure of $K$ and $\theta \in \overline K$ a root of $f$, the rupture field $K(\theta)$ is isomorphic to the field $K(x)/(f)$ and if $\varphi$ is such isomorphism, $\varphi(\theta) = \overline x$ (the class of $x$).

Coming back to your example, we have $K = \mathbb Q$, $\overline K= \mathbb C$ and $f(x) = x^4+1$.

If I understand well, what you name the non symbolic adjunction is $\mathbb Q(\theta)$ and the symbolic one $\mathbb Q(x)/(f)$. Working in one or the other is equivalent. It just means working on one side or the other of the isomorphism $\varphi$. At the end the important topic is that $\theta^4 +1 = 0$ in $\mathbb Q(\theta)$ and ${\overline x}^4+1 = 0$ in $\mathbb Q(x)/(f)$... where in last equality $\overline x$ stands for the class of $x$ in $\mathbb Q(x)/(f)$!

Now, the question you're asked to solve is: factor the quotient $f/(x-\theta)$ in $\mathbb Q(\theta)$. To be coherent with the question and to avoid manipulating equivalence classes, I think it is easier to work in what you call the non symbolic adjunction $\mathbb Q(\theta)$.

Knowing that $f(x)=x^4+1 =(x-\theta)(x + \theta)p(x)$ where $p(x) = x^2+\theta^2$, the problem boils down to decide whether $p$ is irreducible over $\mathbb Q(\theta)$ or have another root in that field. In $\mathbb C$, the set of roots of $f$ is $S=\{\theta, -\theta, i\theta, -i \theta\}$ as $i$ is a primitive fourth root of unity.

You'll verify that both $i \theta$ and $-i\theta$ are roots of $p$ as $\left( \pm i \theta\right)^2=-1 \theta^2=i^2 \theta^2 = -\theta^2$. Hence $f$ factors as $$f(x) = (x-\theta)(x+\theta)(x-i\theta)(x+i\theta)$$ in $\mathbb Q(\theta)$. As an additional bonus, you gain that $i^2 = \theta^4$ and therefore $i$ is either $\theta^2$ or $-\theta^2$. Which can also be retrieved as all roots of $f$ in $\mathbb C$ are eighth root of unity as mentioned in other responses. Therefore we have the factorization in prime factors

$$f(x) = (x-\theta)(x+\theta)(x-\theta^3)(x+\theta^3)$$ of $f$ in $\mathbb Q(\theta)$.

Notes: an important thing to notice, and that I used above, is that $-\theta$ is also a root of $f$ and therefore of $f/(x-\theta)$.

For the fun, let's work in the so called symbolic adjunction $L = \mathbb Q(x)/(f)$.

In $L$, you have $f(\overline x) = {\overline x}^4+1 = 0$. I omit the bars over the elements of $\mathbb Q$, which I shouldn't...

Also in $L(y)$:

$$y^4+1 = (y-{\overline x})(y+{\overline x})(y^2 + {\overline x}^2)$$ and in $L$

$$\left({\overline x}^3\right)^2 + {\overline x}^2= {\overline x}^6 +{\overline x}^2= {\overline x}^4 {\overline x}^2 +{\overline x}^2= -{\overline x}^2+{\overline x}^2=0$$

You retrieve the fact that ${\overline x}^3$ is a root of $p(y) = y^2 +{\overline x}^2$. Finally, the factorization of $f(y)$ in $L(y)$ in prime factors is:

$$f(y) = (y-{\overline x})(y+{\overline x})(y-{\overline x}^3)(y+{\overline x}^3)$$

Here we use nonsymbolic adjunction and state some facts that can be verified.

We set

$\tag 1 \displaystyle \theta = e^{\frac{\pi i}{4}}$

since $\theta^4 + 1 = 0$.

The set

$\tag 2 \Bbb Q [\theta] = \{ f(\theta) \mid f(x) \text{ is a polynomial over } \Bbb Q \}$

is closed under the operations of addition and multiplication defined on $\Bbb C$.

The algebraic structure $(\Bbb Q [\theta], + , \times)$ is a field.

There is a natural field inclusion mapping, ${\displaystyle \iota :\Bbb Q \hookrightarrow \Bbb Q [\theta]}$.

The polynomial $f(x) = x^4 + 1$ can be completely factored into linear factors in the field $\Bbb Q [\theta]$,

$\tag 3 f(x) = (x - \theta) (x - \theta^3)(x - \theta^5)(x - \theta^7)$