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Why doesn't there exist a ring homomorphism between $\mathbf{Q}[x]/(x^2-2) $ and $\mathbf{Q}[x]/(x^2 -3) $?

I see both rings are in fact fields as the polynomials are irreducible, further I know for $T$ to be a ring homomorphism then

$T(1)=1$

$T(x+y)=T(x)+T(y)$

$T(xy)=T(x)T(y)$

I tried proving by contradiction but wasn't really sure how to start

Olórin
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The Problem
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  • To finish Robert's answer it suffices to show that $,\sqrt{2}\not\in\Bbb Q(\sqrt{3}),,$ which follows by this Lemma, which reduces it to $,\sqrt{2},\sqrt{3},\sqrt{6},$ are all irrational. – Bill Dubuque Mar 08 '15 at 18:04

1 Answers1

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In your first ring $2$ has a square root (namely, the class of $x$ modulo $x^2 -2$), and in the second it has not square-root (check it). As any mophism of ring from the first to the second ring would send a square root of $2$ to a square of $2$, no such morphism exists.

Olórin
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  • Why would any mophism of ring from the first to the second ring would send a square root of 2 to a square of 2? – The Problem Mar 08 '15 at 17:02
  • Take $f$ such a morphism. Note $a$ the class of $x$. Then $2 = f(2) = f(a^2) = f(a)f(a) = (f(a))^2$. (Note that for any ring morphism you have $f(n)=n\cdot 1$ when $n\in\mathbf{Z}$.) – Olórin Mar 08 '15 at 17:03
  • Sorry what do you mean by class? and why is 2=f(2)? or have we taken such an f that 2=f(2) – The Problem Mar 08 '15 at 17:07
  • You have a "canonical" morphism $\mathbf{Q}[x]\to \mathbf{Q}[x]/(x^2-2)$, and the class of an element $P\in \mathbf{Q}[x]$ is by definition the image of $P$ by this morphism. Are you familiar with quotients of a ring by an ideal of the ring ? – Olórin Mar 08 '15 at 17:13
  • I know that quotients of a ring by an ideal of the ring also forms a ring and that there's a homomorphism between a ring and this quotient – The Problem Mar 08 '15 at 17:22
  • Then you know everything you need. Do you understand my answer now ? – Olórin Mar 08 '15 at 17:23
  • Yeah I see it now, thanks for helping me once again :-) – The Problem Mar 08 '15 at 17:30
  • @learnmore Read my answer and my first comment to it. – Olórin Mar 08 '15 at 17:49
  • @learnmore Because the square of $\sqrt{3}$ is $3$ and not $2$. Read my answer and my first comment to it. – Olórin Mar 08 '15 at 17:59
  • To finish this answer see my comment on the question. – Bill Dubuque Mar 08 '15 at 18:09
  • @TheProblem it is customary to assume that f(1)=1 for a ring homomorphism between rings with identity. under this assumption we can state that f(2)=f(1+1)=f(1)+f(1)=1+1=2. Otherwise, it is not true. Between any two rings, there is always a homomorphism, namely the trivial one. in that case f(2)=0 – Lozenges Mar 09 '15 at 05:24