I was reading the Atiyah-Macdonald p. 121:

Example. Let $A$ be polynomial ring $k[x_1,\dots,x_n]$ localized at the maximal ideal $\mathfrak{m}=(x_1,\dots,x_n)$.

Then $G_{\mathfrak{m}}(A)$ is a polynomial ring in $n$ indeterminates and so its Poincare series is $(1-t)^{-n}$. Hence we deduce that $\dim A_{\mathfrak{m}}=n$.

I don't know how to show that

$G_{\mathfrak{m}}(A)$ is a polynomial ring in $n$ indeterminates.

I think this is true for $A=k[x_1,\dots,x_n]$. But here $A$ is the localization $k[x_1,\dots,x_n]_{(x_1,\dots,x_n)}$.

Also, I think "$A_{\mathfrak{m}}$" might be a typo since $A$ is already the localization ring.

Any answer and hints are welcome! Thanks a lot.

Here the associated graded ring $G_{\mathfrak{m}}(A)$ is defined by $$G_{\mathfrak{m}}(A)=\bigoplus_{n=0}^{\infty}\mathfrak{m^n}/\mathfrak{m}^{n+1}=(A/\mathfrak{m})\oplus (\mathfrak{m}/\mathfrak{m}^2)\oplus \cdots.$$


In general, if $B$ is a ring with a maximal ideal $\mathfrak{m}$, then $G_\mathfrak{m}(B)\cong G_{\mathfrak{m}}(A)$ where $A$ is the localization $B_{\mathfrak{m}}$. Indeed, the modules $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ are all already $\mathfrak{m}$-local (i.e., all elements of $A\setminus\mathfrak{m}$ act invertibly on them) since they are $A/\mathfrak{m}$-modules, and so localizing doesn't change them. So your case, you can compute $G_\mathfrak{m}(A)$ using the polynomial ring $B=k[x_1,\dots,x_n]$, in which case it is easy to identify $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ as the homogeneous polynomials of degree $n$ and so $G_\mathfrak{m}(B)$ will just be isomorphic to $B$ with its natural grading by degree.

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