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Find the splitting field $E$ of the following polynomials and the degree of the extension

1) $X^4-1\in\mathbb Q[X]$

$X^4-1=(X-1)(X+1)(X^2+1)=(X+1)(X-1)(X+i)(X-i)$ therefore $E=\mathbb Q(i)\cong \mathbb Q[X]/(X^2+1)$ and thus $[E:\mathbb Q]=2$.

I had 0/10 on the question, and the remark was: "the justification is not correct". Could someone explain me why ?

2) $X^7-1\in\mathbb Q[X]$

We have that $$X^7-1=(X-1)(X^6+...+1)$$ with $X^6+...+1$ irreducible.

$$X^7-1=0\iff X^7=1\iff X=e^{\frac{2ik\pi}{7}},\ k=0,...,6.$$ Therefore $$E=\mathbb Q(1,e^{\frac{2i\pi}{7}},...,e^{\frac{12i\pi}{7}})\cong\mathbb Q[X]/(x^6+...+1),$$

and thus $[E:\mathbb Q]=6$.

Is my justification correct ?

3) $X^p-t\in\mathbb F_p(t)[X]$

$t$ is irreducible on $\mathbb F_p(t)$ therefore, by Eisenstein criterion, $X^p-t$ is irreducible over $F_p(t)[X]$.

I have to describe $E$, but how can I solve $X^p=t$ on $\mathbb F_p(t)$ ? I would say that $$X^p=t\iff X=e^{\frac{2ik\pi}{p}}\sqrt[p]t, k=0,...,p-1$$ but the $e^{\frac{2ik\pi}{p}}\sqrt[p]t\notin\mathbb F_p(t)$, therefore I don't know how to do.

4) The minimal polynomials of $\sqrt 2+\sqrt 3$

The minimal polynomial is given by $X^4-10X^2+1$. The roots of this polynomials are $$\sqrt 2+\sqrt 3,\quad -\sqrt 2-\sqrt 3,\quad \sqrt 2-\sqrt 3\quad\text{and}\quad \sqrt 3-\sqrt 2.$$ We have that $$E=\mathbb Q(\sqrt 2+\sqrt 3,-\sqrt 2-\sqrt 3,-\sqrt 2+\sqrt 3,\sqrt 2-\sqrt 3).$$

But $$-\sqrt{2}-\sqrt 3=-(\sqrt 2+\sqrt 3)\in\mathbb Q(sqrt 2+\sqrt 3)$$ $$\sqrt 2-\sqrt 3=\frac{-1}{\sqrt 2+\sqrt 3}\in\mathbb Q(sqrt 2+\sqrt 3)$$ $$-\sqrt 2+\sqrt 3=\frac{1}{\sqrt 2+\sqrt 3}\in\mathbb Q(sqrt 2+\sqrt 3),$$

therefore $$E=\mathbb Q(\sqrt 2+\sqrt 3)\cong\mathbb Q[X]/(X^4-10X+1)$$ and thus $[E:\mathbb Q]=4$.

is it correct ?

idm
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1 Answers1

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For (1): perhaps the lecturer (or whoever marked your exam) wanted some explanation: why does $\;\Bbb Q(i)\;$ is the splitting field of $\;x^4-1\in\Bbb Q[x]\;$ ? Does it contain all the roots of the polynomial? Is this the minimal such field? You only wrote "therefore..." .

For (2): what did you need for the first part? The real thing, imo, is

$$\;x^7=1\iff x=e^{2\pi k i/7}\;,\;\;k=0,1,...,6\;$$

I'd expect though that you'll write, with perhaps a little explanation, that

$$e^{2\pi ki/t}=\left(e^{2\pi i/7}\right)^k\implies E=\Bbb Q(e^{2\pi i/7})$$

Also, why do you say $\;x^6+x^5+\ldots+x+1\;$ is irreducible (over the rationals)? Did you propose a proof for this, or at least mentioned something that could justify this?

For (3): your use of Eisenstein is correct, though it could probably be better to connect it to the "natural" integral domain $\;\Bbb F_p[t]\;$ of which $\;\Bbb F_p(t)\;$ is the fraction field.

Now, you forgot that in characteristic $\;p>0\;$ the Freshman's dream is true: $\;(a+b)^p=a^p+b^p\;$ , so in your case

$$x^p-t=(x-t^{1/p})^p\;,\;\;\text{with}\;\;t^{1/p}\in\overline{\Bbb F_p} =\;\text{some alg. closure of}\;\;\Bbb F_p$$

For (4): why is that the minimal pol. of $\;\sqrt2+\sqrt3\;$ ? Some explanation...?

The last part is nice and correct, imo.

Timbuc
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  • 2) I know that a polynomial of the form $x^{p-1}+...+1$ with $p$ prime is irreducible. 3) therefore $$E=\mathbb Q(t^{\frac{1}{p}})\cong \mathbb F_p(t)[X]/(X^p-t)$$ and thus $[E:\mathbb F_p(t)]=p$. Is the justification correct ? 4) It has been proved in a previous exercise. – idm Dec 24 '14 at 11:50
  • (3) It looks fine, but be careful: don't mix between $;\Bbb Q;$ and $;\Bbb F_p;$ – Timbuc Dec 24 '14 at 12:12