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

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.

• Hint for a far simpler approach: count idempotents. – rschwieb Feb 12 at 11:52

## 1 Answer

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 $$R\to R$$

• Sir but $(0\times R)\cong to (0\times 0\times R)$ so why we can not conclude – SRJ Feb 12 at 8:02
• 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. – Christoph Feb 12 at 8:06
• 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 – SRJ Feb 12 at 8:09
• – SRJ Feb 12 at 8:10
• It doesn't matter, see my edit. – Christoph Feb 12 at 8:20