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Let the ring $S \cong \mathbb Z \times \mathbb Z$ and $I= \{(x,0): x\text{ is in }\mathbb Z\}$ then every $(x,y)$ in $\mathbb Z \times \mathbb Z$ can be written as $(0,y) + (x,0)$ which is an element of $(\mathbb Z\times \mathbb Z)/I$. Then $S\cong I$. ( or can say $S/I$ is isomorphic to $S$). Am I right here?

I have messed up with the concept of quotient ring here. Actually the union of the elements belonging to each of the congruent classes in the quotient ring will obviously be equal to the main ring. I went wrong there.

Sourav Chakraborty
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2 Answers2

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It looks like you're misunderstanding the quotient ring construction when you write

$(0,y)+(x,0)$ which is an element of $(\mathbb Z\times\mathbb Z)/I$

The elements of the quotient rings are equivalence classes (also known as cosets here), so a typical element would be the set $$ \{\ldots,(-1,42),(0,42),(1,42),(2,42),(3,42),\ldots\} $$ consisting of all elements of $\mathbb Z\times \mathbb Z$ that differ from $(1234,42)$ only by something of the form $(x,0)$.

Something that can be written as $(0,y)+(x,0)$ is an element of $\mathbb Z\times\mathbb Z$, not a set of such elements; therefore it is not an element of the quotient ring.

hmakholm left over Monica
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  • you are spot on and just now i realized that. horrible. i am sorry for this. If possible please delete this question of mine and reduce the reputation please. – Sourav Chakraborty Dec 26 '13 at 19:42
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    @SouravChakraborty: I don't see the question as deletion-worthy. I don't really understand the downvotes; in fact you should be praised for explaining enough of your reasoning to make it possible to identify your mistake. And at $+2-2$ you're still 6 rep points ahead. :-) – hmakholm left over Monica Dec 26 '13 at 19:46
  • I agree. It seems that http://math.stackexchange.com/questions/605398/mathbbz-times-mathbbz-is-a-pid-or-not is not needed anymore, perhaps. – Dietrich Burde Dec 26 '13 at 19:51
  • That was the first question I asked in this forum. Please remove which ever you feel unnecessary. – Sourav Chakraborty Dec 26 '13 at 19:54
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No. The ring $\mathbb{Z}\times \mathbb{Z}$ is not an integral domain, since $(1,0)\cdot (0,1)=(0,0)$, so that we have zero-divisors. On the other hand, $S/I\simeq \mathbb{Z}$ is an integral domain. Hence $S$ and $S/I$ cannot be isomorphic rings.

hmakholm left over Monica
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Dietrich Burde
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