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In this post we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:

  • Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

In know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.

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    Consider the square of the element you are extending by and a relation with the multiplicative identity, also recall the defn of field Isomorphisms.2010-11-06
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    An (overkill) way to prove this might be to look at the structure of the Galois group of the field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$.2010-11-06
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    @muad: how can you even determine the structure of that Galois group if you do not know whether $\mathbb Q(\sqrt2)$ is or not equal to $\mathbb Q(\sqrt2)$? (Notice that since they are normal extensions of $\mathbb Q$, they are isomorphic iff they are equal as subfields of $\overline{\mathbb Q}$)2010-11-06
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    @Mariano, do you mean that it can't be done?2010-11-06
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    @muad: I mean that to compute the Galois group you need to know at least whether the two fields are the same or not, and that knowing *that* is enough to answer the original question. Therefore computing the Galois group and understanding its structure is not a useful thing to do.2010-11-06
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    I thought computing the Galois Group was decidable.2010-11-06
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    @muad: Mariano's point is that in order to go ahead and compute the Galois group you must *first* determine whether the two fields are the same. Since the only question here is whether they are the same, you are advocating traveling from point A to point B by first going from A to B, then from B to a very far away point C, and then returning from C to B. It will certainly result in a path that begins at A end ends at B, but...2010-11-06
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    Is $Q(\sqrt{2})=\{a+b \sqrt{2} | a,b \in Q \}$ ??2015-05-05
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    I've come to notice that when I invoke Galois theory on these sorts of questions I usually fall victim to circular reasoning...2017-11-15

6 Answers 6

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More generally: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$.

They are both isomorphic as $\mathbb{Q}$-vector spaces, since they are both of dimension $2$; or more explicitly, every element of $F_1$ can be written uniquely as $a+b\sqrt{d}$ with $a,b\in\mathbb{Q}$ (unique because $\sqrt{d}\notin\mathbb{Q}$), and every element of $F_2$ can be written uniquely as $x+y\sqrt{d'}$ with $x,y\in\mathbb{Q}$. The map $f\colon F_1\to F_2$ given by $f(a+b\sqrt{d}) = a + b\sqrt{d'}$ is additive and $\mathbb{Q}$-homogeneous, clearly bijective, so $F_1$ and $F_2$ are isomorphic as vector spaces over $\mathbb{Q}$.

However, they are never isomorphic as fields; clearly, $d$ is a square in $F_1$. I claim $d$ can only be a square in $F_2$ if $d=d'$. Indeed, if$(x+y\sqrt{d'})^2 = d$. That means that $x^2 + d'y^2 + 2xy\sqrt{d'} = d$, hence $2xy = 0$ and $x^2+d'y^2=d$. If $x=0$, then $d=d'y^2$, so clearing denominators you get $da^2 = d'b^2$ for some $a,b\in\mathbb{Z}$, $\gcd(a,b)=1$; since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$, so $d=d'$. If $y=0$, then $d=x^2$, so $d$ is the square of a rational, contradicting the fact that it is a squarefree integer different from $1$. Thus, of $d$ is a square in $F_2$, then $d=d'$. Hence, if $F_1\cong F_2$, then $d=d'$ (converse is immediate).

Now, since every quadratic extension of $\mathbb{Q}$ is equal to $\mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$ different from $1$, you conclude that any two quadratic extensions are either identical or not isomorphic as fields.

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    @Arturo: $\mathbb{Q}$ homogeneous means?2010-11-06
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    @Arturo: You generalized it nicely. Actually i didnt think of squarefree integers i thought of primes, that is $\mathbb{Q}(\sqrt{p}) \not\cong \mathbb{Q}{\sqrt{q}}$, where $p,q$ are primes!2010-11-06
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    @Chandru1: "Homogeneous" means that $f(ax) = af(x)$ for all scalars $a$. $F$-homogeneous means that it is homogeneous relative to scalars that are in $F$. Here, $\mathbb{Q}$-homogeneous means that if $q\in\mathbb{Q}$ and $\mathbf{x}\in\mathbb{Q}(\sqrt{d})$, then $f(q\mathbf{x}) = qf(\mathbf{x})$. A function between two vector spaces over the field $F$ is linear if and only if it is additive and $F$-homogeneous.2010-11-06
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    @Chandru1: primes are squarefree integers. It is standard to look at squarefree integers; this is one of the first baby steps in algebraic number theory: classify all quadratic extensions of $\mathbb{Q}$.2010-11-06
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    @Arturo: Yes sir! Agreed! But as a student, when i saw this question, the first thing that caught my eye was $2$ and $3$, and what are they, they are primes! I didnt see them as square free integers. Perhaps, with experience, i shall see more.2010-11-06
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    Q(sqrt d) and Q(sqrt d') are isomorphic iff d/d' in Q^2, which is obvious if you exploit the innate symmetry and compare discriminants, cf. my post.2010-11-06
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To prove this: Suspect the fields are not isomorphic, then we can attempt to find a property which holds inside one and does not the other - but whose truth is preserved by isomorphism.

In the field $\mathbb{Q}(\sqrt{2})$ there is an element which satisfies the field property $x^2=2$. There is no element in $\mathbb{Q}(\sqrt{3})$ which satisfies this, but suppose for a contradiction that there was an isomorphism $\psi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ we would have $\psi(x^2) = \psi(2)$ which is equivalent to $\psi(x)^2 = \psi(1)+\psi(1)$ and since $\psi(1) = 1$ we have an element of $\mathbb{Q}(\sqrt{3})$ which, squared, is 2.

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    @muad: Thanks a lot muad!2010-11-06
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    Oh this was very easy..2010-11-06
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    @Muad: Your answer is far from complete because you have omitted the proof of the most essential point, viz. that 2 is not a square in Q(sqrt(3)). Giving that proof amounts to doing what I said.2010-11-06
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    @muad: How do you know that $\mathbb{Q}(\sqrt{3})$ does not contain an element whose square is $2$? (It's true, but what argument are you giving for it?)2010-11-06
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    $(a+b\sqrt{3})^{2} = 2 \Longrightarrow (a^{2}+3b^{2}-2) + 2ab\sqrt{3} = 0 \Longrightarrow A+B\sqrt{3}=0$, where $A$ and $B$ are rationals.doesn't that give u a contradiction2010-11-06
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    @Chandru1: Yes, that will work. You should see it through to the end: since $\sqrt{3}$ is irrational, $1$ and $\sqrt{3}$ are linearly independent over $\mathbb{Q}$, so your last equation implies $A = B = 0$, and so forth. Why not write this up nicely as an answer?2010-11-06
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    @Chandru1: You need to prove more, e.g. why doesn't the same "proof" work to also (falsely) show that 3 isn't a square in Q(sqrt(3)) ? In particular you need B nonzero to get a contradiction.2010-11-06
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    (Actually, Arturo has already written up the argument that I suggested to Chandru1 to complete, very nicely and in more generality.)2010-11-06
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HINT $\ $ Compare discriminants: $\rm\displaystyle\ \big\{\big(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt 2)\big\}\: =\ 2\ \mathbb Q^{\:2}\ \ $ vs. $\rm\ \ 3\ \mathbb Q^{\:2}\ $ for $\rm\ \mathbb Q(\sqrt 3)\ $

Note that if $\rm\ \alpha,\: \alpha'\ \not\in\mathbb Q\ $ are conjugate then they remain so under any field isomorphism since their minimal polynomial $\rm\ (x-\alpha)\ (x-\alpha')\ $ is in $\rm\:\mathbb Q[x]\:$ so it is fixed by any isomorphism.

In fact quadratic fields are characterized uniquely be their discriminant.

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    You probably should explain what you mean by $\mid$, as $3$ is plainly a unit in $\mathbb Q(\sqrt 3)$.2010-11-06
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    @Mariano: Thanks, I've edited it for clarification.2010-11-06
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    Here $\alpha = \sqrt{2}$ and $\alpha' = \sqrt{3}$? So you are computing $\left(\frac{\sqrt{3}-\sqrt{2}}{2} \right)^2$? It boils down to the fact that $\text{irr}(\sqrt{2}, \mathbb{Q}) \neq \text{irr}(\sqrt{3}, \mathbb{Q})$?2010-11-06
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    @Trevor: No, they are conjugate elements in one of the quadratic fields.2010-11-06
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Short answer: the splitting fields of $x^2-2$ and $x^2-3$ over $\mathbb{Q}$ cannot be the same field or isomorphic fields. For instance, there are infinite odd primes $p$ such that $\left(\frac{2}{p}\right)=1$ and $\left(\frac{3}{p}\right)=-1$ (any prime $\equiv 17\pmod{24}$, for instance) and infinite odd primes $p$ such that $\left(\frac{2}{p}\right)=-1$ and $\left(\frac{3}{p}\right)=1$ (any prime $\equiv13\pmod{24}$, for instance). In particular $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a quadratic extension of both $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ and these fields cannot be isomorphic.

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The only possibility for an isomorphism of the extension fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$, fixing $\mathbb{Q}$ (the prime subfield) would be $\varphi:\sqrt{2}\mapsto \sqrt{3}$. But $\varphi$ preserves products and so:

$2 = \varphi(\sqrt{2}\sqrt{2}) = \varphi(\sqrt{2})\varphi(\sqrt{2}) = \sqrt{3}\sqrt{3} = 3$ contradiction. So the aforementioned fields can't be isomorphic.

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I'm not sure if this is correct, but can't we also use the fact that if $F_{1}$ and $F_{2}$, as splitting fields, have the same minimal polynomial $f(x) \in K[x]$ then there exists an isomorphism between them.

So basically you just have to show that they don't have the same roots and that obviously $x^{2}-2$ and $x^{2}-3$ aren't the same.