- #76

fresh_42

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##f(x)##I present to you the most mediocre and non-repeatable solution to #10

since I'm writing this on my phone, I'm going to write ##a## instead of ##\alpha##.

First we observe that ##f(x)## has no repeated roots, as ##f'(x)=3x^2-3## which has roots ##\pm 1##, which are not roots of ##x##.

Furthermore, the rational root test says any rational root of ##f## must be a factor of the constant term divided by a factor of the highest degree term, hence any rational root must be of the form ##\pm 1## which again is not a root.

**Could you elaborate this a bit? What is the rational root test?**If ##f(x)=(x-\alpha)(x-\beta)(x-\gamma)##, why can't we have ##\alpha=2,\beta=\sqrt{2},\gamma=1/2\sqrt{2}##? It is not hard to do it explicitly, but if you use a theorem, can you quote it?

This is all correct, except a tiny gap. You must rule out the case ##\alpha=2-\alpha^2##.So ##f## is a separable polynomial, and is irreducible since any factoring must include a polynomial of degree 1 and hence a root.

Therefore the automorphisms of the splitting field correspond to permutations of the roots of ##f##. If ##a## a root implies ##2-a^2## is a root (to be shown below), ##\mathbb{Q}(a)## has at least two and hence all three of the roots of ##f(x)## since the roots can be factored out leaving only a linear factor left. This shows ##\mathbb{Q}(a)## is the splitting field for ##f##. ##[\mathbb{Q}(a):\mathbb{Q}]=3##, so the group of automorphisms has magnitude 3. Every permutation on three elements either swaps two and leaves the last unchanged, or cycles all three. Since the order of the permutation must divide the order of the group, the single swaps are not possible, and the Galois group must be ##\mathbb{Z}_3##, with the automorphisms mapping roots ##a_1##, ##a_2## and ##a_3## as either ##a_1\to a_2 \to a_3 \to a_1## or the inverse.

I have never heard about the term magnitude in this context. Do you have a reference? Just out of curiosity.

A shortcut after observing ##f(\alpha)=f(2-\alpha^2)=0## and ##\alpha\neq 2-\alpha^2## would have been to note, that we have a Galois extension and thus a transitive automorphism group.From here, it suffices to show that if ##a## is a root of ##f(x)##, so is ##2-a^2##, and that this does not fix a single root while swapping the other two. If it swapped two roots, we would have ##2-(2-a^2)^2=a##. Expanding the left hand side and moving ##a## over yields

$$2-4+a^4-4a^2-a=0.$$

$$a^4-4a^2-a-2=0.$$

We know ##a^3=3a+1## and hence ##a^4=3a^2+a## so this gives us

$$-a^2-2=0.$$

But we know that ##a## cannot satisfy a polynomial of degree 2 in ##\mathbb{Q}## since ##[\mathbb{Q}(a):\mathbb{Q}]=3## and by the tower lemma there are no intermediate fields of degree 2 that ##\mathbb{Q}(a)## can contain

So all we need to do now is show if ##f(a)=0##, that ##2-a^2## is also a root. This is just algebra:

$$f(2-a^2) = (2-a^2)^3 - 3(2-a^2) -1$$

Expanding gives

$$=8-12a^2 + 6a^4 - a^6 -6+3a^2-1$$

$$=-a^6+6a^4-9a^2+1$$

We use ##a^3=3a+1## to get ##a^6=3a^4+a^3## and hence

$$=3a^4-a^3-9a^2+1$$

Similarly, ##a^4=3a^2+a## so we get

$$=-a^3+3a+1=-f(a)=0$$

Hence the automorphism ##\sigma## exists as desired.

Depends on the definition ofTo solve for ##\sqrt{12-3a^2}## I did dumb things. Namely, I guessed it was going to involve a ##2-a^2## and just tried to get there. I used wolfram alpha to get ##a##, noticed ##\sqrt{12-3a^2}## was a little more than twice ##2-a^2## when ##a## was picked to be the smallest root in magnitude, and then saw adding ##a/2## got me the rest of the way there. Doing a bit of algebra gives the claim:

$$\sqrt{12-3a^2}=4+a-2a^2.$$

First we show that

$$12-3a^2=(4+a-2a^2)^2.$$

Expanding the right hand side yields

$$16+a^2+4a^4+8a-16a^2-4a^3.$$

Moving the ##12-3a^2## over and simplifying terms we need to show

$$0=4+8a-12a^2-4a^3+4a^4.$$

Dividing by 4 leaves us proving

$$0=1+2a-3a^2-a^3+a^4.$$

Plugging in our trusty formula of ##a^4= 3a^2+a##, this reduces to

$$0=1+3a-a^3=-f(a)=0.$$

So we have the squared formula. Taking square roots, we now need to only show that ##4+a-2a^2>0##. As an aside, we know all the roots are real numbers since complex roots will come in conjugate pairs and then conjugating the roots yields an automorphism that isn't cycling all three roots, so this is actually either a positive or negative number, as are all the roots.

it suffices to show that there exists at least one root with magnitude <1, as then ##|4+a-2a^2| > 4 -|a| - 2|a|^2## and ##|a|,|a|^2 < 1## in this case. But the roots have to be in between the turning points of the polynomial, which we computed to be ##\pm 1## earlier. In particular there is one root smaller than -1, one root larger than 1, and one in between as desired. So ##4+a-2a^2 >0## as required.

You must have had a better idea in mind for how to come up with the polynomial expression for ##\sqrt{12-3a^2}## than my dumb way of doing it.

*dumb*. I made a long division by ##x-\alpha##, observed (via the extrema you already calculated) ##\alpha\approx 1/3## which thus must be the closest to ##0##, and got the desired radical for ##2-\alpha^2=(x-\varphi \pm \rho)##.

Long division results in something with ##\sqrt{\alpha}##, so there is a way to come from one to the other expression. The fact, that they were identical, shortcuts the process a lot.Edit:I realized you can also just say if there is a polynomial, it must be at most quadratic since you can use the cubic polynomial relation to reduce the degree. Then you just need to find ##p(a)^2=12-3a^2## and use the cubic relation again to reduce the degree of the left hand side. It's still not clear to me how you would have known this is even contained in the field from first principles though.