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

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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What have you tried? –  Mariano Suárez-Alvarez Nov 6 '10 at 18:26
Consider the square of the element you are extending by and a relation with the multiplicative identity, also recall the defn of field Isomorphisms. –  BBischof Nov 6 '10 at 18:26
@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 $\overline\mathbb Q$) –  Mariano Suárez-Alvarez Nov 6 '10 at 20:49
@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. –  Mariano Suárez-Alvarez Nov 6 '10 at 22:06
@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... –  Arturo Magidin Nov 6 '10 at 23:44

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? –  anonymous Nov 6 '10 at 19:36
@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! –  anonymous Nov 6 '10 at 19:39
@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. –  Arturo Magidin Nov 6 '10 at 19:40
@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 $\mathbb{Q}$. –  Arturo Magidin Nov 6 '10 at 19:42
@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. –  anonymous Nov 6 '10 at 19:43

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! –  anonymous Nov 6 '10 at 19:07
@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. –  Bill Dubuque Nov 6 '10 at 19:15
$(a+b\sqrt{3})^{2} = 2 \Longrightarrow (a^{2}+3b^{2}-2) + 2ab\sqrt{3} = 0 \Longrightarrow A+B\sqrt{3}=0$, where $A$ and $B$ are rationals.doesn't that give u a contradiction –  anonymous Nov 6 '10 at 19:34
@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? –  Pete L. Clark Nov 6 '10 at 19:39
(Actually, Arturo has already written up the argument that I suggested to Chandru1 to complete, very nicely and in more generality.) –  Pete L. Clark Nov 6 '10 at 19:45

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 $\mathbb Q(\sqrt 3)$. –  Mariano Suárez-Alvarez Nov 6 '10 at 19:00
@Mariano: Thanks, I've edited it for clarification. –  Bill Dubuque Nov 6 '10 at 19:30
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 $\text{irr}(\sqrt{2}, \mathbb{Q}) \neq \text{irr}(\sqrt{3}, \mathbb{Q})$? –  PEV Nov 6 '10 at 19:40
@Trevor: No, they are conjugate elements in one of the quadratic fields. –  Bill Dubuque Nov 6 '10 at 19:50