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

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

I 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.

Sidharth Ghoshal
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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. – BBischof Nov 06 '10 at 18:26
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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 $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$. –  Nov 06 '10 at 18:40
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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 $\overline{\mathbb Q}$) – Mariano Suárez-Álvarez Nov 06 '10 at 20:49
  • @Mariano, do you mean that it can't be done? –  Nov 06 '10 at 21:23
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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. – Mariano Suárez-Álvarez Nov 06 '10 at 22:06
  • I thought computing the Galois Group was decidable. –  Nov 06 '10 at 22:23
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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... – Arturo Magidin Nov 06 '10 at 23:44
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    Is $Q(\sqrt{2})={a+b \sqrt{2} | a,b \in Q }$ ?? – Bozo Vulicevic May 05 '15 at 09:36
  • I've come to notice that when I invoke Galois theory on these sorts of questions I usually fall victim to circular reasoning... – Tob Ernack Nov 15 '17 at 16:18

5 Answers5

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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.

Arturo Magidin
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    @Arturo: $\mathbb{Q}$ homogeneous means? –  Nov 06 '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! –  Nov 06 '10 at 19:39
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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. – Arturo Magidin Nov 06 '10 at 19:40
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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 $\mathbb{Q}$. – Arturo Magidin Nov 06 '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. –  Nov 06 '10 at 19:43
  • 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. – Bill Dubuque Nov 06 '10 at 22:21
  • @ArturoMagidin It may seem very easy question for you but can you please tell me , if you have some spare time that why $Q(\sqrt{d})$=$Q(\sqrt{d'})$ implies d =d' . Pardon me , but i am a beginner in Rings and field and so unablle to deduce this !! –  Aug 09 '20 at 19:40
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    @user795826 You don’t bother to say what exactly is tripping you, which makes answering the question difficult. If they were equal, then there is some $a+b\sqrt{d}$ whose square is $d’$. This requires $a^2+d b^2 = d’$ and $2ab=0$. Since $d$ and $d’$ are assumed to be square free, this easily yields that you must have $a=0$, $b=\pm 1$, and $d=d’$. – Arturo Magidin Aug 09 '20 at 19:46
  • @user795826 And now I see that I made that very same argument in the post ten years ago. Did you read the post before asking? – Arturo Magidin Aug 09 '20 at 20:47
  • @arturo Magidin, can't we say something more generally:two simple extensions are isomorphic iff they have been obtained through adjoining roots of two distinct irreducible polynomials – Ibrahim Islam Oct 15 '20 at 09:13
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    @Ibs: No; here you have two simple extensions obtained by adjoining roots of two distinct irreducible polynomials, and they are not isomorphic. And adjoining the roots of $x^2-2$ and of $(x-1)^2-2$, both irreducible, distinct, yield the exact same extension, hence they are isomorphic. – Arturo Magidin Oct 15 '20 at 11:40
  • @Arturo, sorry I did mistake in writing my doubt,I once again I put my question:prove or disprove that two simple extensions are isomorphic iff they have been obtained through adjoining roots of an irreducible polynomial – Ibrahim Islam Oct 15 '20 at 13:11
  • @Ibs: That’s still wrong. Every simple extension is obtained by adjoining the root of an irreducible polynomial; you can have simple extensions of different degrees, which cannot be isomorphic. over $\mathbb{Q}$, every finite extension is simple. I don’t know what you are trying to ask, but. the answer is almost certainly going to be “no”. – Arturo Magidin Oct 15 '20 at 13:58
  • @arturo can you give me counter example in which two non isomorphic simple extensions are obtained through adjoining roots of same irreducible polynomial.I think you didn't understand my question well,......I meant same irreducible polynomial in my question – Ibrahim Islam Oct 15 '20 at 14:54
  • @Ibs: It’s not that I didn’t understand your question; it’s that what you asked is not what you meant. You did not say “same irreducible”, but ”an irreducible”. It’s not an if and only if statement, as I already noted: the extensions you get by adjoining a root of $x^2-2$ and the extension you get by adjoining a root of $x^2-2x-1$ are equal. Of course, if $p(x)$ is an irreducible, and $\alpha$ is a root of $p(x)$ in some algebraic closure, then $F[\alpha]\cong F[x]/p(x)F[x]$ regardless of the root. Any two roots yield isomorphic extensions. – Arturo Magidin Oct 15 '20 at 14:59
  • @Ibs: That is not just well known, but a standard result on extensions. The converse fails, though, spectacularly. – Arturo Magidin Oct 15 '20 at 15:00
  • @arturo,thanks,Now my confusion gone – Ibrahim Islam Oct 15 '20 at 15:17
  • Let us continue this discussion in chat. – Ibrahim Islam Oct 15 '20 at 15:36
  • @Ibs: No; if you still have questions, then make a post and ask your question. This is way too long for a comment thread on a 10 year old question, and I have no interest or desire to go chat. – Arturo Magidin Oct 15 '20 at 15:38
  • @Arturo:Sorry it was done mistakenly,I have no doubt regarding this question.... – Ibrahim Islam Oct 15 '20 at 16:08
  • Dear @ArturoMagidin why is it true that: $\textit{since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$}$? – Barreto Dec 09 '22 at 05:52
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    @Sarah $da^2 = d'b^2$, so $b^2$ divides $da^2$. Since $\gcd(a,b)=1$, then $b^2$ divides $d$. Since $d$ is squarefree, the only square that divides it is $1$. – Arturo Magidin Dec 09 '22 at 15:30
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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.

  • Oh this was very easy.. –  Nov 06 '10 at 19:09
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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. – Bill Dubuque Nov 06 '10 at 19:15
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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?) – Pete L. Clark Nov 06 '10 at 19:17
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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 contradiction –  Nov 06 '10 at 19:34
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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? – Pete L. Clark Nov 06 '10 at 19:39
  • @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. – Bill Dubuque Nov 06 '10 at 19:41
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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.) – Pete L. Clark Nov 06 '10 at 19:45
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Hint $\ $ Compare discriminants: $\{\:\!(\alpha-\alpha'\big)^2:\ \alpha \in \mathbb Q(\sqrt 2)\:\!\} =\,2\:\!\mathbb Q^{2}\, $ vs. $\, 3\:\!\Bbb 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.

Generally quadratic fields are characterized uniquely by their discriminant. The point of expressing the proof in this way is to highlight this fundamental invariant of number fields.

Bill Dubuque
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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.

Jack D'Aurizio
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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.

ks_evizr
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    How do you conclude that the map would have to send $\sqrt{2}$ to $\sqrt{3}$? The fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{2}+1)$ are both quadratic, and even though the map fixing $\mathbb{Q}$ and sending $\sqrt{2}$ to $\sqrt{2}+1$ is not a field isomorphism, there is a field isomorphism between them. Your claim that any potential morphism between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ would have to send $\sqrt{2}$ to $\sqrt{3}$ is unfounded. – Arturo Magidin Feb 23 '19 at 08:09