Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.