# False proof that $R_1[x]\cong R_2[x]$ implies $R_1\cong R_2$ (which is not true)

As it can be seen in [1], there exist commutative rings (with $$1$$) such that $$R_1[x]\cong R_2[x]$$ but $$R_1$$ and $$R_2$$ are not isomorphic.

If we have one single ring $$R$$, then it is clear that $$R\hookrightarrow R[x]\twoheadrightarrow R,$$ in which the first arrow is the natural inclusion and the second arrow is evaluation in $$0$$, is the identity of $$R$$ and so is an isomorphism.

It seems that one could generalize this to our problem by considering the following morphism $$R_1\hookrightarrow R_1[x]\overset{\sim}{\to}R_2[x]\twoheadrightarrow R_2,$$ in which the middle arrow is our given isomorphism. This is clearly a ring homomorphism and intuitively it should be an isomorphism with $$R_2\hookrightarrow R_2[x]\overset{\sim}{\to}R_1[x]\twoheadrightarrow R_1$$ as inverse. Why doesn't this work?

• Did you work out what happens if you try your idea on Hochster's example? The problem (and the whole reason the claim is false) is that the isomorphism between $R_1[x]$ and $R_2[x]$ doesn't neatly map elements of $R_1$ to $R_2$. Apr 22, 2020 at 8:12
• @Magdiragdag Frankly, I can't say that I understand it very well. There are some stuff there that I am not comfortable with yet. Apr 22, 2020 at 8:19

Consider the ring $$R_1 = R_2 = \mathbb Z[t]$$. Now, instead of the identity map between $$R_1[x] = \mathbb Z[t,x]$$ and $$R_2[x] = \mathbb Z[t,x]$$, look at the map that switches $$x$$ and $$t$$: $$f(x,t) \mapsto f(t,x)$$. This map is an isomorphism, but your composition $$R_1\hookrightarrow R_1[x]\overset{\sim}{\to}R_2[x]\twoheadrightarrow R_2,$$ is not; it maps $$t$$ to $$0$$.
So, even in the case where there is an isomorphism between $$R_1$$ and $$R_2$$, your construction does not necessarily give one.