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We say that an ideal $\alpha$ of $A$ is finitely generated if $\alpha =(x_1,\cdots,x_n)=\sum_{i=1}^{n} Ax_i$, i.e. finitely generated as an $A$-module.

Then how we call if $\alpha$ is generated by all the finite products of the $x_i$? In other words, every element of $\alpha$ is a polynomial in $A[x_1,\cdots,x_n]$ with no constant term. It is similar to the finitely generated $A$-algebra, but it is not an $A$-algebra since $\alpha$ is not a ring and does not contain constant term.

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E.g., $Ax^2\subseteq Ax$. – Jonas Meyer Jun 14 '11 at 1:38
@Jonas: I'm not sure I understand your comment. Could you expand a bit? – Zev Chonoles Jun 14 '11 at 1:40
@Zev: Sorry I was unclear. As a consequence of that containment, the ideal generated by $x\in A$ is the same as that generated by $\{x,x^2\}$, and so on for higher powers. Similarly, $Ax_ix_j\subseteq Ax_j$, etc. I didn't mean to be cryptic, but I was brief because you had already answered the question. – Jonas Meyer Jun 14 '11 at 1:43
up vote 5 down vote accepted

If $\alpha$ is generated by all the finite products of the $x_i$, then it is generated by the $x_i$. In other words, $$(x_1,x_2,\ldots,x_n)=(x_1,x_2,\ldots,x_n,x_1^2,x_1x_2,\ldots,x_n^2,\ldots).$$ So there is not a separate concept of an ideal being finitely generated like there is for $A$-modules vs. $A$-algebras.

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Thanks. Actually I knew this for some times and I forgot this again. Now I will not forget this. – Gobi Jun 14 '11 at 1:49

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