I am asking this because while reading about Galois theory I came across this:

Let $K$ be a Galois extension of $F$. Let $\lambda$ map $K$ onto $\lambda K$ be an isomorphism. Then $\lambda K$ is a Galois extension of $\lambda F$. Indeed, if $K$ is the splitting field of the polynomial $f$, then $\lambda K$ is the splitting field of $\lambda F$.

While working on this I have tried to come up with an example (and for various other theorems). However, I cant seem to find an isomorphism between $\mathbb Q[\sqrt2]$ and another field. I can prove that it cant be isomorphic to another quadratic extension but thats it. Does an isomorphic field even exist? If not or $\mathbb Q[\sqrt2]$ can you provide me an example where a splitting field is isomorphic to another splitting field that is not identical? Thank you

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    $\begingroup$ It's like if you asked "are there other sets than $\{\varnothing\}$ which have cardinality one?" $\endgroup$ – Pierre-Yves Gaillard Jul 6 '17 at 18:46
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    $\begingroup$ Do you mean, "which other fields $\mathbb{Q}(\sqrt{m})$ are isomorphic to $\mathbb{Q}[\sqrt{2}]$ other than $\mathbb{Q}[\sqrt{2}]$ itself?" $\endgroup$ – Dietrich Burde Jul 6 '17 at 18:48
  • $\begingroup$ No i don't mean what other fields Q(sort(m)) as I said in my question: "I can prove that it cant be isomorphic to another quadratic extension". I just want an exmple for the theorem above and for later ones that deal with isomorphisms between galois groups and fields. $\endgroup$ – Daniele1234 Jul 6 '17 at 18:58

I'm guessing that you might really be interested in the following.

Lemma. If $L$ is any field, then it has at most one subfield $K$ that is isomorphic to $\Bbb{Q}(\sqrt2)$.

Proof. Unless $L$ has characteristic zero you are not going to find such a subfield, so we can assume that. Then the prime field of $L$ is isomorphic to $\Bbb{Q}$ (by a unique isomorphism). Therefore we can without loss of generality assume that $\Bbb{Q}\subseteq L$. To get to the required $K$ we need the polynomial $x^2-2$ to have zeros in $L$. For if $a^2=2$ then $K=\Bbb{Q}(a)$ works. But, $x^2-2$ is quadratic so it cannot have more than two zeros in $L$. And if $a$ is one root then $-a$ is the other. But $\Bbb{Q}(a)=\Bbb{Q}(-a)$, so the subfield is unique (as a set). QED.

This is in sharp contrast to $\Bbb{Q}(\root3\of2)$. Adjoining a single root of $x^3-2$ does not necessarily give the others free of charge. Indeed, we have three distinct subfields of $\Bbb{C}$ that are all isomorphic to $\Bbb{Q}(\root3\of2)$.

However, if $a$ generates a cyclic cubic extension then essentially the same argument works. So for example $M=\Bbb{Q}(\cos\dfrac{2\pi}7)$ is the only subfield of $\Bbb{C}$ isomorphic to $M$. To see this let $m(x)=x^3+x^2-2x-1$ be the minimal polynomial of $2\cos(2\pi/7)$. If $m(a_1)=0$ for some $a_1\in\Bbb{C}$ then it is easy to show that $a_2=a_1^2-2$ is another. And $a_3=a_2^2-2=-a_1^2-a_1+1$ is the third. This allows us to conclude that $\Bbb{Q}(a_1)$ contains all the zeros of $m(x)$, namely $2\cos(2\pi/7), 2\cos(4\pi/7)$ and $2\cos(8\pi/7)$. Those are then necessarily $a_1,a_2,a_3$ in some order. Hence $\Bbb{Q}(a_1)=M$.

The same holds for all (finite) Galois extensions of $\Bbb{Q}$. Each and every one of them is isomorphic to a unique subfield of $\Bbb{C}$.

For such uniqueness results to work it is necessary to fix a big ambient field $L$ and work inside it. I used $L=\Bbb{C}$. Otherwise we have "incomparable" fields like the one constructed in Hagen von Eitzen's answer.

In algebra texts splitting fields are often introduced without a big ambient field. There are several reasons for this: A) The teacher doesn't need to worry about the students having seen a proof of the FTA. B) The text will soon use the "uniqueness up to isomorphism" variant of the construction of a splitting field to prove the uniqueness (up to isomorphism) of finite fields of a given cardinality, and this is a very convenient way of doing that. Observe that when splitting polynomials over $\Bbb{Z}_p$ we don't have a suitable algebraically closed ambient field available!


If you define $\Bbb Q[\sqrt 2]$ as the smallest subfield of $\Bbb C$ that contains $\sqrt 2$, then here's a totally different field that is isomorphic to it: $\Bbb Q[X]/(X^2-2)$, one (of two possible) isomorphism being given by $f(X)+(X^2-2)\Bbb Q[X]\mapsto f(\sqrt 2)$.


For square free integers $n$ and $m$ we have that $$\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m}) \Longleftrightarrow n=m,$$ see here. This implies that $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{\ell})$ if and only if $\ell=2k^2$.

  • $\begingroup$ I know this hence why I said in my question: "I can prove that it cant be isomorphic to another quadratic extension but thats it". I know that it cant be isomorphic to another extension Q(sqrt(n)). $\endgroup$ – Daniele1234 Jul 6 '17 at 18:57
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    $\begingroup$ Well, it can be isomorphic to another quadratic extension: $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{8})$. $\endgroup$ – Dietrich Burde Jul 6 '17 at 18:58
  • $\begingroup$ But 2 does not equal 8 how can this be the case? $\endgroup$ – Daniele1234 Jul 6 '17 at 19:00
  • $\begingroup$ How do you arrive at the conclusion: "this implies that..." $\endgroup$ – Daniele1234 Jul 6 '17 at 19:01
  • $\begingroup$ Because $8=2k^2$ for $k=2$. The implication comes from the fact, that then $\ell$ cannot be square-free - so it must be $2$-times a square. $\endgroup$ – Dietrich Burde Jul 6 '17 at 19:01

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