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Let $A$ be a semilocal ring with Jacobson radical $m$ and let $I$ be an ideal of definition, i.e. an ideal such that $m^{\nu} \subset I \subset m$. Consider the associated graded ring of $A$, given by $gr_I(A)=\bigoplus_{n \ge 0} I^n/I^{n+1}$. Suppose that $I=(x_1,\cdots,x_s)$. Then $gr_I(A)=A/I[\bar{x}_1,\cdots,\bar{x}_s]$, where $\bar{x}_i$ is the image of $x_i$ in $I/I^2$. Certainly, we have that $A/I[\bar{x}_1,\cdots,\bar{x}_s] \cong A/I [X_1,\cdots,X_s] / J$, where $X_i$ are free and $J$ is some ideal.

Question: Why is $J$ homogeneous?

Remark: $J$ is the kernel of the map $A/I[X_1,\cdots,X_s] \rightarrow A/I[\bar{x}_1,\cdots,\bar{x}_s]$ given by $X_i \mapsto \bar{x}_i$ and so $J$ is the ideal of relations between the $\bar{x}_1,\cdots,\bar{x}_s$ over $A/I$. Why do these relations need to be generated by homogeneous elements?

Edit: This appears in a more specific setting in the proof of Theorem 14.4 of Matsumura's Commutative Ring Theory. It goes like that: Let $(A,m,k)$ be a $d-$dimensional regular local ring. Then $m$ is generated by $d$ elements and $gr_m(A)=k[X_1,\cdots,X_d]/I$, where $I$ is a homogeneous ideal.

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Kernels in the category of graded modules are created by the forgetful functor to modules. That is, you just compute kernels pointwise. Now conclude. –  Martin Brandenburg Jun 16 '13 at 19:18
    
@MartinBrandenburg: Got it. –  Manos Jun 16 '13 at 19:23
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