Let $R$ be a commutative ring which is a subring of a commutative ring $A$ which in turn is a subring of $R[x,y]$ (all rings and subrings are with unity). Then is it true that there is an ideal $J$ of $A$ such that $A/J \cong R$ ?

(The question is motivated from the fact that the claim holds when $A=R[x,y]$, in particular.)

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    $\begingroup$ Do you ask that the composition of the inclusions $R\to A$ and $A\to R[x,y]$ is the natural inclusion $R\to R[x,y]$? $\endgroup$ Aug 30, 2016 at 14:10

2 Answers 2


Yes: let $\phi: A \longrightarrow R$ be the composition of the inclusion $A \subseteq R[x,y]$ with the quotient map $R[x,y] \longrightarrow R$. Since the composition of $\phi$ with the inclusion $R \subseteq A$ is the identity on $R$, the map $\phi:A \to R$ is surjective. It follows that you can take $J=A \cap I$ where $I$ is the kernel of the quotient map $R[x,y] \longrightarrow R$.


The map $\alpha:R[x,y]\to R[x,y]$ defined by $\alpha(f)=f(0,0)$ is a homomorphism of rings that fixes $R$, and indeed $\alpha(f)\in R$ for all $f$. The kernel $K$ is easy to compute; it is the set of all $f\in R[x,y]$ with zero constant term. The restriction of a homomorphism to a subring is always a homomorphism, so this gives us a homomorphism $\alpha:A\to R$ that fixes $R$. The kernel is $J=K\cap A$, the set of all polynomials contained in $A$ with zero constant term, and this is the ideal we seek.

Not only is there an ideal such that $A/J\simeq R$, but there is a homomorphism $\epsilon:A\to A$ such that $\epsilon(f)\in R$ for all $f\in A$ and if $r\in R$ then $\epsilon(r)=r$. Thus $A$ retracts onto $R$, and $\epsilon$ is a homomorphism of $R$-algebras.


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