12
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ In this case, I think it's important to note that $$x^4+1=\frac{x^8-1}{x^4-1}$$In other words, the roots of the polynomial are the non-trivial roots of unity of $8$. So, since roots of unity are generated by a non-trivial root of unity, we can call any of one the roots of $x^4+1$, for example, $\frac1{\sqrt2}(1+i)$, $\zeta_8$, and then note that $$\mathbb Q(\zeta_8)=\mathbb Q[x]\;/\;(x^4+1)$$ $\endgroup$ – Don Thousand Feb 10 at 21:41
  • 2
    $\begingroup$ Is there a reason you allow yourself to write $\sqrt{i}$ but not $\sqrt{-i}$? $\endgroup$ – Captain Lama Feb 10 at 21:50
  • $\begingroup$ @CaptainLama, I better did so indeed, since I've written the same things twice ($\sqrt i$ and $\frac{√2}{2}(1+i)$). Thank you $\endgroup$ – master_clown Feb 11 at 8:00
2
$\begingroup$

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:

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

$\endgroup$
2
$\begingroup$

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 $\Bbb Q [\theta]$,

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

$\endgroup$
1
$\begingroup$

I think it's important to realize what's taken as obvious here and what not.

Obvious: there is a root in some extension, and there is a prime factorization in another extension

Non obvious: all roots are equivalent in the sense that:

Non obvious: the polynomial can be factored in the simple field extension $Q(\theta)$ for any root $\theta$

In particular, the question is probably supposed to be attacked using a more algebraic manner than simply writing down the factorization over $\mathbb{C}$. Knowing the factorization over $\mathbb{C}$ helps figuring out the algebra at play however.

$\endgroup$
1
$\begingroup$

Firstly observe that $x^8-1=(x^4+1)(x^4-1)$. Now if $\alpha$ is a primitive 8th root of unity, then $\alpha$ is a root of $x^4+1$. Also, other roots of $x^4+1$ are $\alpha^3,\alpha^5,\alpha^7$.( For instance, you can choose $\alpha= \frac{1}{\sqrt2}+i \frac{1}{\sqrt2}$).

So adjoining $\alpha$ to $\mathbb Q$, $(\mathbb Q(\alpha))$ is the splitting field of $x^4+1$. It is easy to see that $\alpha^3,\alpha^5,\alpha^7 \in \mathbb Q(\alpha)$.

So prime factorization of $x^4+1$ in $\mathbb Q(\alpha)$ is $(x-\alpha)(x-\alpha^3)(x-\alpha^5)(x-\alpha^7)$.

So basically what your intuition is correct for the first part (nonsymbolic adjunction). i.e firstly we choose what is the algebraic closure, and then we factorize the given polynomial and then we adjoin a root. This course is kinda pre deterministic since we have already determined the algebraic closure.


On the other hand (symbolic adjunction), assume you don't know anything about complex numbers. So we don't know about the algebraic closure of $\mathbb Q$. But what interesting is still we can construct a filed $K$ which contains a root of $x^4+1$ and $K$ contains an isomorphic copy of $\mathbb Q$.

Explicitly, $K=\mathbb Q[x]/<x^4+1> = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb Q \}$

(Since $x^4+1$ is irreducible in $\mathbb Q[x]$, $K$ is a field)

Now if $\pi$ is the canonical projection of $\mathbb Q[x]$ to $K$, then the image of $\pi$ restricted to $\mathbb Q$ gives the isomorphic copy of $\mathbb Q$. With that, if we identify $\mathbb Q$ as a subfield of $K$

Now observe this: if we let $\overline x=\pi(x)$,

$\pi(x^4+1)=\pi(x^4)+\pi(1)=(\pi(x))^4+\pi(1)=\overline x^4+\pi(1)$

since $\pi(x^4+1)=0$, This implies that $\overline x$ is a root of $x^4+1$ in $K$.

What is interesting is $K \cong \mathbb Q(\alpha)$

$\endgroup$
  • $\begingroup$ Gune, special thanks for explaining what $\mathbb Q[x]/(x^4 + 1)$ is.That was a problem for me $\endgroup$ – master_clown Feb 13 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.