In other words, if $f$ is a ring homomorphism from $\mathbb{Q}[\sqrt{2}]$ to $\mathbb{Q}[\sqrt{3}]$, then $f(r) = 0$ for all $r \in \mathbb{Q}[\sqrt{2}]$...

I was trying to see if the assumed $f$ contradicts any one of axioms of ring homomorphisms, but didn’t see one. I was thinking using the fact that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are fields so the only ideals are trivial ideals. Since $\ker(\mathbb{Q}[\sqrt{2}])$ is also an ideal, I want to show that this $\ker(\mathbb{Q}[\sqrt{2}])$ is not the zero ideal. Then I didn’t go too far because I have no other information about the structure of these two fields. Also, I was thinking about representing any non-zero $r$ in $\mathbb{Q}[\sqrt{2}]$ and to map it and somehow hopefully use the no zero-divisor property of Integral Domain to show that $r$ is mapped to $0$…But I’m not sure if the specific $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ can give us stronger hypothesis…


Hint: If $\phi$ is a homomorphism then $\phi(\sqrt 2)^2=\phi(2)$. If $\phi$ is non-trivial then $\phi(2)=2$...

Oops. It's been pointed we need to show that $\phi(1)=1$ if $\phi$ is nontrivial. In any case, if $x=\phi(1)$ then $x^2=x$; since there are no zero divisors in $\Bbb Q[\sqrt3]$ this shows that $\phi(1)$ is $0$ or $1$, and if $\phi(1)=0$ then $\phi$ is trivial.

  • $\begingroup$ You need to reference why that second statement is true. For example, the field $F=\mathbb Q[\sqrt{2}]$ has non-trivial ring homomorphisms $F\to R$ for some $R$ with $\phi(2)\neq 1+1$ in $R$. (We are, apparently, not specifically restricting to homomorphisms which send $1$ to $1$, since the trivial homomorphism does not.) $\endgroup$ – Thomas Andrews Jan 27 '16 at 16:04
  • $\begingroup$ @ThomasAndrews Oops, thanks. Gotta go to class, will think about it later... $\endgroup$ – David C. Ullrich Jan 27 '16 at 16:13
  • $\begingroup$ @ThomasAndrews No hints please - got it, no time to write it down... $\endgroup$ – David C. Ullrich Jan 27 '16 at 16:17
  • $\begingroup$ Yeah! But this only tells us this "one" instance of $\phi$ is not well-defined. How do we argue that all of them except for the trivial one are not well-defined? $\endgroup$ – nekodesu Jan 27 '16 at 16:19
  • $\begingroup$ @Shiyue Huh? (it was a hint, by the way...) $\endgroup$ – David C. Ullrich Jan 27 '16 at 20:49

Hint: think of what the square of the image of $\sqrt2$ should be.


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