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Can anyone help me with these two doubts of mine:

  1. Is the ring $\mathbb R\oplus \mathbb R$ an Integral Domain or a Division Ring?

  2. My notes state that the ring of Gaussian integers(i.e. $\mathbb Z[i]=\{a+bi~~\big|a,b\in \mathbb Z\}$) is not a field although it is an integral Domain. I don't understand which element in this ring does not have a multiplicative inverse.

Kindly help with these two doubts of mine. Thanks in advance for any help.

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  • $\begingroup$ How do you define multiplication on the structure? $\endgroup$ – quid Oct 19 '14 at 13:51
  • $\begingroup$ @quid I define multiplication in $\mathbb R \oplus \mathbb R$ as usual multiplication under dot product... $\endgroup$ – kittuu Oct 19 '14 at 13:56
  • $\begingroup$ Well $\mathbb{Z}[i]$ is an Euclidean ring with norm $N(a+ib)=a^{2}+b^{2}$. And the necessary and sufficient condition that $a\in\mathbb{Z}[i]$ is a unit is $N(a)=N(1)=1$. So the elements in $\mathbb{Z}[i]$ which have multiplicative inverse are precisely those $(a,b)\in\mathbb{Z}$ which satisfy $a^{2}+b^{2}=1$. $\endgroup$ – crskhr Oct 19 '14 at 13:56
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For the second question:

  1. this can be considered as a subset of the complex numbers.

  2. recall that the inverse must be in the ring too (not just some complex number that would exist).

  3. observe that the square of the absolute value of every number in the ring is an integer, it is $a^2+ b^2$. Therefore, an element can be only invertible if its absolute value is $1$. Every other element cannot be invertible.

For the first, it depends on how you define multiplication: if you define it coordinate wise, that is $(a,b)(c,d)= (ac,bd)$ which is what I assume you mean (though this is not really the dot product which would be $ac+bd$ and not yield a composition law at all), then you have zero-divisors for example $(3,0)(0,9)=(0,0)$ and the structure thus is neither an integral domain nor a division ring.

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  • $\begingroup$ just make me clear with 1 thing why is $\mathbb R\oplus \mathbb R $ not a division ring from your explanation.. $\endgroup$ – kittuu Oct 19 '14 at 14:06
  • $\begingroup$ A division ring cannot have any zero-divisors, as a zero-divisor cannot be invertible. But you might prefer a direct argument: it is not a division ring since for example $(2,0)(a,b)= (2a,0)$ and thus $(2,0)$ is not invertible as $(2,0)(a,b)\neq (1,1)$ for each $(a,b)$. $\endgroup$ – quid Oct 19 '14 at 14:09
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1) $\mathbb R\oplus \mathbb R$ is not a Domain, since $(1,0)$ is a divisor of zero, so it can't be either a Division Ring.

2) I think 2 hasn't an inverse..

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