# Rees ring associated to an ideal

I am reading Atiyah-Macdonald, the chapter on completions.

Let $A$ be a ring (not graded), and let $\mathfrak{a}$ be an ideal of $A$. Then we can form a graded ring $A^*=\oplus_{n=0}^{\infty}\mathfrak{a}^n$. The book then says that if $\mathfrak{a}$ is finitely generated, say by $x_1,\dots,x_r$, then $A^*=A[x_1,\dots, x_r]$. How do we get this?

Is it because of the following equivalence? For a graded ring $B$, $B$ is noetherian if and only if $B_0$ is noetherian and $B$ is finitely generated as a $B_0$ algebra? But I am still not able to precisely write down why $A^*=A[x_1,\dots, x_r]$. Any help will be appreciated!

• Be awared that there are relations between the $x_i$'s in $A[x_1,\cdots, x_r]$. – Cantlog Feb 19 '14 at 20:51
• This is a (bad) notation. – Cantlog Feb 19 '14 at 22:00
• @Cantlog Was the Noetherian condition removed deliberately? – user1729 Feb 20 '14 at 15:26
• @user121097 I don't know much about ring theory, it was just that someone removed it in an edit. Which is why I asked. – user1729 Feb 20 '14 at 18:09
• It is not me who removed the noetherian condition. But this condition is actually unnecessary: assuming $\mathfrak a$ is finitely generated is enough. – Cantlog Feb 21 '14 at 9:00

Actually the best way to understand the Rees ring is to define it as a subring of a polynomial ring $A[t]$, that is, $A^*=\oplus_{n=0}^{\infty}\mathfrak a^nt^n$. When $\mathfrak a=(x_1,\dots,x_r)$ we have $A^*=A[x_1t,\dots,x_rt]$ and this is easily shown since every element of $\mathfrak a^n$ is a linear combination (with coefficients in $A$) of monomials of degree $n$ in $x_1,\dots,x_r$.