I have a problem with the solution of the tasks of abstract algebra.Help please.

$F=({\bf Q};+;\cdot),K=({\bf R},+;\cdot)$.

  1. Determine the degree of the extension $F_K^* (\sqrt 2,\sqrt 3):F]$.
  2. Designate a primitive element of the extension $[F_K^* (\sqrt 2, \sqrt[3]3):F]$, over the field $F_K^* (\sqrt 2)$.
  3. Determine the Galois group of the field $F_K^ * (\sqrt 2, \sqrt 3)$ over the field $F$, i.e., the group $(G, \circ)$, where $G$ is the class of all automorphisms $Y$ of the field $F_K^* (\sqrt 2, \sqrt 3)$ such that $Y(x) = x$ for $x\in {\bf Q}$.

Definition. For any Z:Univ;A:Class;F,K:Field, where F⊏K, Z is an extension of the field F of the field K of class A <=> Z = $F_K^*$ (A): = K (F__supp ∪ A | Field) .

Answer for 1. Is my thinking is right? Is this a good answer? Does it solved?

since $[\Bbb Q(√2,√3):\Bbb Q] = [\Bbb Q(\sqrt 2,\sqrt 3):\Bbb Q(\sqrt 2)]*[\Bbb Q(\sqrt2):\Bbb Q]$

and since $x^2 - 2$ is irreducible over $\Bbb Q$, and

$x^2 - 3$ is irreducible over $\Bbb Q(\sqrt 2)$

(the former is obvious, and latter can be shown by assuming there

is $q + r\sqrt 2$ such that $(q+r\sqrt 2)^2 = 3$, for $q,r \in \Bbb Q$.

that is $(q^2 + 2r^2) + (2qr)\sqrt2 = 3$, leading to:

$q^2 + 2r^2 = 3qr = 0$. if $r = 0$, then $q^2 = 3$, impossible since $\sqrt3$ is irrational.

if $q = 0$, then $2r^2 = 3$, writing $r = a/b$ for integers with $gcd(a,b) = 1$,

we have $2a^2 = 3b^2$. if $p$ is a prime dividing $b$, then $p$ divides $2$

(since $p$ does not divide $a$, and thus $a^2$). since $p^2$ divides $3b^2$,

$p^2$ must also divide $2$, and this is impossible. hence $b^2 = 1$,

but there is no integer solution to $2a^2 = 3$, since $2a^2$ is even and $3$ is odd).

therefore $[\Bbb Q(\sqrt2,\sqrt3):Q] = 2*2 = 4$.

I tried to solve the second. Is my solution is correct?

Answer for 2. a primitive element of Q(√2,√3) is √2+√3. it's clear that

√2+√3 lies in Q(√2,√3), hence Q(√2+√3) is contained in Q(√2,√3).

note that (1/2)[(√2 + √3)^3 - 9(√2+√3)] = √2, so

√2 is in Q(√2+√3), hence √3 = (√2+√3) - √2 is as well.

hence they are equal fields.

Answer for 3. consider the polynomial $f(x) = (x^2 - 2)(x^2 - 3)$ in $\Bbb Q[x]$.

any automorphism of $\Bbb Q(\sqrt2,\sqrt3)$ that fixes $\Bbb Q$ must send a root of $f$ to another root of $f$.

furthermore, such an automorphism must send a root of $x^2 - 2$ to a root of $x^2 - 2$,

and similarly for $x^2 - 3$. now $\Bbb Q(\sqrt2,\sqrt3)$ is a normal separable extension of $\Bbb Q$

since $f$ is separable, and $\Bbb Q(\sqrt2,\sqrt3)$ is the splitting field of f over $\Bbb Q$. hence:

$|Gal(\Bbb Q(\sqrt2,\sqrt3)/\Bbb Q)| = 4$.

we see that the following 4 assignments all determine automorphisms:

$\sqrt2\mapsto \sqrt2$, $\ \sqrt3\mapsto \sqrt3$ (the identity)

$\sqrt2\mapsto -\sqrt2$, $\ \sqrt3\mapsto \sqrt3$ (sending only $\sqrt2$ to its conjugate)

$\sqrt2\mapsto \sqrt2$, $\ \sqrt3\mapsto -\sqrt3$ (sending only $\sqrt3$ to its conjugate)

$\sqrt2\mapsto -\sqrt2$, $\ \sqrt3\mapsto -\sqrt3$ (sending both $\sqrt2$ and $\sqrt3$ to their conjugates)

thus this is all of them. since every automorphism has order 2,

this is isomorphic to the klein 4-group, V.

  • $\begingroup$ Could you check whether these arrangements are good? Do not I made a mistake? $\endgroup$ – Marta Jan 24 '13 at 15:43
  • $\begingroup$ My question is whether the solution to the second task is correct? if anyone noticed some error? $\endgroup$ – Marta Jan 24 '13 at 19:43

Yes, all seems correct. What about question 2?

  • $\begingroup$ Aha. I guess, it is only $\sqrt[3]3$, and a similar argument works as in 1. $\endgroup$ – Berci Jan 23 '13 at 14:47
  • $\begingroup$ Yhm $\sqrt[3]3$ . yes $\endgroup$ – Marta Jan 23 '13 at 17:01

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