1
$\begingroup$

For an integral domain $R$, the rings $R\times R $ and $R\times R\times R$ are not isomorphic

My attempt:

On contrary suppose that both are isomorphic then if G is prime ideal of one ring then its isomorphic copy must be prime ideal of other

we will construct projection map $p_1:R\times R\to R$ as $p_1(a,b)=a$

Now here kernel is $R$

then by prime ideal theorem $(R\times R )/R\cong R$ which is integral domain so R is prime ideal of R.

Now we will again construct projection map $p_2:R\times R\times R\to R\times R$

as $p_2(a,b,c)=(a,b)$

Now here kernel is $R$ Now $(R\times R \times R )/R\cong R\times R $ but it is easy to show that $R\times R $ is not integral domain.

So our assumption is wrong .

Hence both are not isomorphic.

Is my argument are valid?

Thanks in advanced.

$\endgroup$
  • 3
    $\begingroup$ Hint for a far simpler approach: count idempotents. $\endgroup$ – rschwieb Feb 12 at 11:52
1
$\begingroup$

Your argument is not valid: the quotients $(R\times R)/(0\times R)$ and $(R\times R\times R)/(0\times 0\times R)$ not being isomorphic does not rule out the existence of an isomorphism $\varphi\colon R\times R\to R\times R\times R$. It just rules out that such an isomorphism can satisfy $\varphi(0\times R)=0\times 0\times R$.


Let me clarify that for two isomorphic rings $R$ and $R'$ with ideals $I\subset R$ and $I'\subset R'$ that are also isomorphic as rings, you can not conclude that $R/I$ and $R'/I'$ are isomorphic rings:

Take any non-zero ring $S$ and let $R=\bigoplus_{i=1}^\infty S$ with component-wise addition and multiplication. Let $R'=R$, $I=R$ and $$ I' = \left\{ \, (0,s_1,s_2,\dots)\in R \,\middle|\, s_i\in S\,\right\}. $$ Note that the shift map $$ (s_1,s_2,\dots) \mapsto (0,s_1,s_2,\dots) $$ is a ring isomorphism from $I$ to $I'$. However, $R/I=0$ while $R/I'\cong S$.

Also note that the shift map is no longer surjective when extended to a map $R\to R$

$\endgroup$
  • $\begingroup$ Sir but $ (0\times R)\cong to (0\times 0\times R)$ so why we can not conclude $\endgroup$ – SRJ Feb 12 at 8:02
  • $\begingroup$ Note that $(R\times R)/R$ doesn't even make sense since $R$ is not a subset of $R\times R$. You have to be more careful about what ideals you are considering. Maybe this example (for groups) helps: $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3\mathbb Z$ are not isomorphic groups although the groups $2\mathbb Z$ and $3\mathbb Z$ are isomorphic. $\endgroup$ – Christoph Feb 12 at 8:06
  • $\begingroup$ Sir $2Z$ and $3Z$ are not ring isomorphic. as if there is such isomorphism (2+2)=2.2 there is no element in 3Z with this property $\endgroup$ – SRJ Feb 12 at 8:09
  • $\begingroup$ See math.stackexchange.com/questions/1277844/… $\endgroup$ – SRJ Feb 12 at 8:10
  • 1
    $\begingroup$ It doesn't matter, see my edit. $\endgroup$ – Christoph Feb 12 at 8:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.