isomorphic quotient rings of polynomial ring and Hilbert functions
Let $k$ be a field, $R=k[x_1,\cdots,x_n]$ and $I,J$ homogeneous ideals of $R$. Denote by $H_I(s), H_J(s)$ the Hilbert functions of $I,J$ respectively. If $R/I, R/J$ are isomorphic as graded rings, then $H_I(s) = H_J(s), \forall s \ge 0$. Intuitively, it seems to me that the converse is not true, i.e. we might have that $H_I = H_J$, yet $R/I, R/J$ are not isomorphic as graded rings. Am i right? Any counterexamples?
The way i think about it is like that: if $H_J = H_I$, then for every degree $s \ge 0$ we can find $k$-vector space isomorphism $f_i: R_s/I_s \rightarrow R_s/J_s$. With these $f_i$, we can define a $k$-vector space isomorphism $f:R/I \rightarrow R/J$. My concern is that this need not be a homomorphism of $R$-modules. Any comments regarding this morphism?
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$\begingroup$Here's an idea, which might be wrong.
Let $I=(x_1^2)$ and $J=(x_1x_2)$ in $R=k[x_1,x_2]$. Then the gradings for $R/I$ and $R/J$ certainly agree at $s=0,1$. At $s=2$, we have generators $\{x_1x_2,x_2^2\}$ for $R/I$ and $\{x_1^2,x_2^2\}$ for $R/J$, so they have the same dimension. At $s=3$, we have $\{x_1x_2^2,x_2^3\}$ for $R/I$ and $\{x_1^3,x_2^3\}$ for $R/J$. This should generalise to all higher gradings as well, so these have the same Hilbert function.
However, these aren't isomorphic because in $(R/I)_1$, there is an element $a$ (i.e. $x_1$) such that $a^2=0$. But in $(R/J)_1$, any nontrivial element squared is nontrivial. Therefore these aren't isomorphic.
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