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.

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

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 $R\to R$

  • $\begingroup$ 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

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