I am currently working to find all ring homomorphisms $\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})$.

My work so far: Since $\mathbb{Q}(\sqrt{2})$ is a field extension of $\mathbb{Q}$, a field, it is obviously a field as well. Therefore, its only ideals are $\{0\}$ and itself. Since every homomorphism's kernel is an ideal, we have that either $Ker(\phi)=\{0\}$ or $\mathbb{Q}(\sqrt{2})$.

In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective).

My question: I think it is true that in the former case $\phi(1)=1$, but am not sure why. If this is the case, can I prove inductively that $\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})$?

Thank you in advance for your help.

  • $\begingroup$ I know very little ring theory, so I only have some comments... We know that if phi(1)=0, then phi(a) = phi(1*a) = phi(1)*phi(a) = 0 (for all a), hence the trivial homomorphism. By similar logic, it must be otherwise that phi(1) = 1 as we expect: phi(1*a)=phi(a). Note: phi(1*a) = phi(1)*phi(a)= RHS = phi(a). Hence phi(1) = 1. Secondly, I don't see how this implies that phi(a) = a must be the only possible remaining homomorphism. Take phi(a) = a^(-1). I.e. phi (x) = x^(-1). What do you think? $\endgroup$ – Just_a_fool Apr 8 '17 at 23:02
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    $\begingroup$ This is a duplicate of math.stackexchange.com/questions/2204425/… (which was answered in comments). $\endgroup$ – Rob Arthan Apr 8 '17 at 23:10

Hint: Let $\varphi \colon \mathbb{Q} (\sqrt{2} \,) \to \mathbb{Q} (\sqrt{2} \, )$ be a nonzero ring homomorphism. It suffices to know the values $\varphi(1)$ and $\varphi(\sqrt{2}\,)$ (why?). We must have $\varphi(r) = r$ for every $r \in \mathbb{Q}$, so $\varphi(1) = 1$. Also, $$ 2 = \varphi(2) = \varphi(\sqrt{2} \sqrt{2} \,) = \varphi(\sqrt{2} \,) \varphi(\sqrt{2} \,) = [\varphi(\sqrt{2} \,)]^2. $$ So what possible ring homomorphisms could $\varphi$ be?

Note: I have used the fact that $\varphi$ must act as the identity on $\mathbb{Q}$. Make sure you understand why this is so (another exercise).

  • $\begingroup$ Thank you Nunya, this answer is informative, but left some for me to figure out. That is, every element of $\mathbb{Q}(\sqrt{2})$ can be expressed as integer linear combinations of $1$ and $\sqrt{2}$. Since $\phi$ is a ring homomorphism, we can explicitly show that $\phi (a+b\sqrt{2})=a+b\sqrt{2}$. Hence, $\phi (r)=r, \forall r \in \mathbb{Q} (\sqrt{2})$ $\endgroup$ – user322548 Apr 10 '17 at 14:55
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    $\begingroup$ @EthanZell Every element of $\mathbb{Q}(\sqrt{2}\,)$ can be written as a linear combination of $1$ and $\sqrt{2}$ with rational coefficients. So for every $x \in \mathbb{Q}(\sqrt{2} \,)$, there exist $a$, $b \in \mathbb{Q}$ such that $x = a + b \sqrt{2}$. $\endgroup$ – Mark Twain Apr 10 '17 at 19:50
  • $\begingroup$ Thank you for the clarification! $\endgroup$ – user322548 Apr 10 '17 at 23:55

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