# The associated graded ring of the localization $k[x_1,\dots,x_n]_{(x_1,\dots,x_n)}$

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.