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  1. $\mathbb{Z}$ and $\mathbb{2Z}$

My solution: To prove they are isomorphic as groups, I take the mapping $f: \mathbb{Z} \rightarrow \mathbb{2Z}$ defined by $f(x)=2x$. I prove it's a homomorphism and surjective and I am done.

To prove they are not isomorphic as rings, I take the equation $x^2=1$. It has solutions $x=1,-1$ in $\mathbb{Z}$ but no solutions in $\mathbb{2Z}$ and are hence not isomorphic.

  1. $\mathbb{Z}[\sqrt2]$ and $\mathbb{Z}[\sqrt5]$

My solution: To prove they are isomorphic as groups, I take the mapping

$f: \mathbb{Z}[\sqrt2] \rightarrow \mathbb{Z}[\sqrt5]$ defined by $f(a+b\sqrt2)=a+b\sqrt5$. I prove it's a homomorphism and surjective and I am done.

Here, to prove they are not isomorphic as rings, I take the equation $x^2=2$, which has solutions $x=\sqrt2, -\sqrt2$ in $\mathbb{Z}[\sqrt2]$ but no solution in $\mathbb{Z}[\sqrt5]$.

Is this the correct approach to proving non-isomorphism as rings? That an equation has a solution in one ring but not in another?

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  • $\begingroup$ Your arguments are correct. $\endgroup$ – Lev Borisov Mar 27 '16 at 14:39
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    $\begingroup$ Can one speak of the equation $x^2=1$ in $2\mathbb{Z}$? $\endgroup$ – sqtrat Mar 27 '16 at 14:40
  • $\begingroup$ Yes, this is the correct approach, see also this duplicate. $\endgroup$ – Dietrich Burde Mar 27 '16 at 14:46
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    $\begingroup$ As sqrat has pointed out, one has to be a little careful about the formulation of the property $P$ that one ring has and the other doesn't. One ring has a unit and the other does not. Or if one wants equations, the equation $x^2=x$ has two solutions in one of the rings, and only one in the other. $\endgroup$ – André Nicolas Mar 27 '16 at 14:50
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By Definition, a ring homomorphism $f: R \rightarrow R'$ must preserve addition and multiplication and must map the multiplicative identity of $R$ to the multiplicative identity of $R'$. In your example, the ring $R'=2\mathbb{Z}$ does not have a multiplicative identity. So the two rings are not isomorphic (there is no isomorphism, or even a homomorphism from one ring to the other, for that matter).

To show that the rings $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{5}]$ are not isomorphic, you can use your idea that $x^2=2$ has no solution in the latter ring. But you need to justify why this method works. Here is a proof. Suppose there is an isomorphism $f$ between these two rings that takes $a+b\sqrt{2}$ to $a'+b'\sqrt{5}$. Since $f$ must take the identity to the identity, $f$ takes 1 to 1' (here, 1' is the identity in the second ring, and actually equals the integer 1; the primes are just to make things clearer). Since $f$ preserves sums, $f$ must take $1+1$ to $1'+1'$. Now, $(0+1 \sqrt{2})(0+1\sqrt{2}) = 1+1$ in the first ring. We can apply $f$ to both sides. Since $f$ preserves sums and products, we get the equation $(x'+y'\sqrt{5})^2 = 2$, where $x'+y'\sqrt{5}$ is the image of $(0+1\sqrt{2})$ under $f$. This equation has no solutions, and so we get a contradiction. Thus, there does not exist an isomorphism $f$ from the first ring to the second.

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