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Theorem: Let $d,d'>1$ be square-free integers. Then $\mathbb{Q}(\sqrt{d})$ and $\mathbb{Q}(\sqrt{d'})$ are isomorphic as fields of and only if $d=d'$.

In this question I saw the following proof:

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 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).

I don't understand this step: "since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$". Why is that true?

Xam
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non-abelian group of order 9
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1 Answers1

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You have $da^2=d'b^2$ and $gcd(a,b)=1$. Then we can conclude that $b^2|d$ via the following lemma

Lemma: Let $a,b,c$ be elements of a UFD such that $gcd(a,b)=1$ and $a|bc$. Then $a|c$.

The idea is that any prime factor of $a$ can't be a prime factor of $b$, and so must be a prime factor of $c$. Since every prime factor of $a$ is a prime factor of $c$, $a|c$.

However, we know that $d$ is square free, so $b=\pm 1$. Similar analysis works for $a$.

Stella Biderman
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