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Let $f(x)=x^4+16 \in \mathbb{Q}[x]$.

  1. Split $f(x)$ into a product of first degree polynomials in $\mathbb{C}[x]$.

  2. Show that $f(x)$ is an irreducible polynomial of $\mathbb{Q}[x]$.

  3. Find the splitting field $E$ of $f(x)$ and the degree of the extension $[E:\mathbb{Q}]$.

I have done the following:

  1. $f(x)=(x^2-4i)(x^2+4i)=(x-2\sqrt{i})(x+2\sqrt{i})(x-2\sqrt[3]{i})(x+2 \sqrt[3]{i})=(x-2 e^{\pi i/4})(x+2 e^{\pi i /4})((x-2 e^{\pi i/6})(x+2e^{\pi i/6})$

Is it correct??

  1. $f(x)$ is irreducible in $\mathbb{Q}$. If it were not irreducible, then it could be written as a product of polynomials of $\mathbb{Q}[x]$ as followed:

    • It can be written as a product of four first degree polynomials: $f(x)=(x-2\sqrt{i})(x+2\sqrt{i})(x-2\sqrt[3]{i})(x+2 \sqrt[3]{i})$ But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.

    • It can be written as a product of two second degree polynomials: $f(x)=(x^2-4i)(x^2+4i)$ But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.

    • It can be written as a product of a first degree and a third degree polynomial: $f(x)=(x-2\sqrt{i})\left [(x+2\sqrt{i})(x^2+4i)\right ] \\ =(x-2\sqrt{i}) (x^3+4ix+2\sqrt{i}x^2+8i\sqrt{i})$ But the coefficients are not in $\mathbb{Q}$, So, it cannot be written in that way.

Is it correct??

  1. The splitting field is $E=\mathbb{Q}(\pm 2 e^{\pi i/4}, \pm 2 e^{\pi i/6})=\mathbb{Q}(e^{\pi i/4}, e^{\pi i/6})$

Is it correct??

$Irr(e^{\pi i/4}, \mathbb{Q})=x^4+1$

$[\mathbb{Q}(e^{\pi i/4}):\mathbb{Q}]=4$

How can I continue to find $[\mathbb{Q}(e^{\pi i/4}, e^{\pi i/6}): \mathbb{Q}]$??

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  • $\begingroup$ You didn't check all possible ways to combine the factors. If you want to do this method, you cannot just try a few combinations. An alternative that happens to work for this polynomial is to use linear substitutions $x = 2y$ and then $y = z+1$ to get to a polynomial where you can use Eisenstein's criterion with prime $2$. $\endgroup$ – user21820 Nov 29 '14 at 11:25
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    $\begingroup$ Note that $x^4 + 16 = x^4 - (-16) = x^4 - (4i)^2 = (x^2 + 4i)(x^2 - 4i)$. $\endgroup$ – N. F. Taussig Nov 29 '14 at 11:28
  • $\begingroup$ @N.F.Taussig Yes, that is what I meant... It is a typo... $\endgroup$ – Mary Star Nov 29 '14 at 11:58
  • $\begingroup$ @user21820 Could you explain me further what I have to do?? $\endgroup$ – Mary Star Nov 29 '14 at 12:01
  • $\begingroup$ What did you get after the two substitutions? And what does Eisenstein's criterion say? $\endgroup$ – user21820 Nov 29 '14 at 12:03
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For $\;k=0,1\;$ :

$$\begin{align*}&\;\;\;\;\,i=e^{\frac\pi2i+2k\pi i}\implies \sqrt i\;\;\;\;=\;e^{\frac\pi4i+k\pi i}=\begin{cases}e^{\frac\pi4i}\;=\;\;\,\frac1{\sqrt2}(1+ i)\\{}\\e^{\frac{5\pi i}4}=-\frac1{\sqrt2}(1+i)\end{cases}\\{}\\ &-i=e^{-\frac\pi2i+2k\pi i}\implies \sqrt{-i}=e^{-\frac\pi4i+k\pi i}=\begin{cases}e^{-\frac\pi4i}=\;\;\,\frac1{\sqrt2}(1-i)\\{}\\e^{\frac{3\pi i}4}\;=-\frac1{\sqrt2}(1-i)\end{cases}\end{align*}$$

From here

$$x^4+16=(x^2-4i)(x^2+4i)=$$

$$=(x-\sqrt2(1+i))(x+\sqrt2(1+i))(x-\sqrt2(1-i))(x+\sqrt2(1-i)$$

If, for example, you multiply the first and third factor above you get

$$x^2-2\sqrt2\,x+4$$

and likewise with the second and fourth factors:

$$x^2+2\sqrt2\,x+4$$

This way, you can get the decomposition of the polynomial over $\;\Bbb R\;$ and check at once that there are no real, and thus also rational, roots.

With this you already have solved $\;(1)-(2)\;$

Now, for $\;(3)\;$ : observe that

$$r=\sqrt2(1+i)\implies r^2=2\cdot2i=4i\implies r^4=-16\implies \;[\Bbb Q(\sqrt2(1+i):\Bbb Q]=4$$

and a possible $\;\Bbb Q$-basis of this linear space is (with$\; w:=\sqrt2(1+i)=2\,e^{\frac\pi4i}$)

$$\left\{\;1\,,\,\,w\,,\,\,w^2=4i\,,\,\,w^3=4\sqrt2(-1+i)=-4\sqrt2(1-i)\;\right\}$$

Check that all the roots of $\;x^4+16\;$ are a linear combination of this basis, and you're done.

Of course, the two lines after "Now, for $\;(3)\;$" are superfluous as we already know that, but whatever.

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