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Let $k$ be a field, $m$ be a positive integer and $R$ be the subring $k[x,xy,xy^2,…,xy^m]$ of the polynomial ring $k[x,y]$. Let $B$ be the quotient ring $R/xR$. Then $B$ is the finitely generated $k$-algebra $k[xy,xy^2,…,xy^m]$.

How do I compute the associate reduced ring for the $k$-algebra $B$? Is this reduced ring always isomorphic to the polynomial ring $k[t]$?

I have verified that this associated reduced ring for $B$ is isomorphic to the polynomial ring $k[t]$ for the cases $m=1,2$. What I did was to find all the nilpotent elements in the $k$-algebra generators for $B$ and take the ideal in $B$ generated by these elements as the nilradical of $B$. Then I took the quotient ring of $B$ by its nilradical to compute the reduced ring associated to $B$. But I am not sure about the general case. It seems to me that in the general case, there are many relations to be considered when the quotient of $B$ by its nilradical is taken. This makes the determination of the isomorphism type of the associated reduced ring of $B$ challenging.

Could someone please help me with this problem? Thank you so much for your kind help.

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    $\begingroup$ "associate reduced ring" = quotient by the nilradical? $\endgroup$ Commented Feb 3, 2023 at 1:46
  • $\begingroup$ Yes, by the associated reduced ring of a commutative ring A, I meant the quotient of A by its nilradical. $\endgroup$
    – Boris
    Commented Feb 3, 2023 at 3:47

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In $R/xR$, all $xy^l, l<m$ are nilpotent. To see this, notice that $(xy^l)^m= x^{m-l}(xy^m)^l$. Thus, modulo nilpotents, $R/xR$ is generated by the single element $xy^m$ over $k$. The rest is clear.

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  • $\begingroup$ Thank you so much for your kind help. Your explanations are very helpful. $\endgroup$
    – Boris
    Commented Feb 3, 2023 at 17:10

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