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Recently, I was confused by two different definitions in atiyah macdonald. In page 30, it says, A ring is said to be finitely generated if it is finitely generated as a $\mathbf Z$ algebra (i.e $A=\mathbf Z[a_1,a_2...a_n]$), but a ring is also an ideal, so it is finitely generated by $(1)$( in ideal is finitely generated sense as usual). Two definition is not the same, for example, $\mathbf Q$ is finitely generated in latter sense, but it can not be finitely generated in the former sense sice there is infinitely many prime numbers. Moreover, I cannot find this definition in the website, Does it wrong?

lee
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    Being finitely generated as a ring and being finitely generated as an ideal over itself (equivalently, as a module over itself) are just two different terms. They are not supposed to be equivalent, they are just different. – Mark Dec 30 '23 at 02:25
  • Comments should only be used to clarify, not answer the question. See How do comments work for more information. – Martin Brandenburg Dec 30 '23 at 02:36
  • Others have pointed out that "finitely generated algebra" and "finitely generated ideal" (a.k.a. "finitely generated module") are different things. I'll just point out that the former is often called "finitely generated" and the latter is often called just "finite", i.e. "finite" in the context of rings means that it is a finitely generated ideal (module) over itself. This is a bit confusing though, because "finite" in this context does not mean "finite as a set". – Dave Dec 30 '23 at 03:11
  • Quite some work has been done to more or less avoid all the confusions caused by the fundamental error to ignore forgetful functors. – Martin Brandenburg Dec 30 '23 at 06:01

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I will give you two answers. The first one is a bit handhavy but is sufficient for mathematical practice. The second is a bit more complicated, and it supports the first one with some mathematical rigor.

Answer 1

When we are talking about a ring which is finitely generated, it is understood that we are refering to the notion of a finitely generated ring. When we are talking about an ideal which is finitely generated, it is understood that we are refering to the notion of a finitely generated ideal. You are correct (almost, see below) that every ring is also an ideal of that ring, but in practice there will be no confusion since in any case we can add "(as an ideal)" or "(as a ring)" after the statement that some object is finitely genrated.

Answer 2

Rings and ideals are not of the same type. I will present here just one way of making this precise (there are others as well). Hence, any property of rings does not hold (in fact, cannot be applied!) for ideals, and vice versa. The statement "The ring $R$ is an ideal of $R$" is not completely correct, even though every mathematician uses it and can work with it without making any mistakes.

The correct statement would be: For every ring $R$ there is an ideal of $R$ that has the same underlying set (or even the same additive group) as $R$.

In fact, remember that, generally speaking, it is important to distinguish between an object and its underlying set (in case this object has an underlying set at all!). Unfortunately, this fact is usually forgotten because when speaking and writing about mathematical objects we usually don't make any such distinction. Let us do it here, and denote the underlying set of, say, an algebraic structure $X$ by $|X|$.

An ideal $I$ is a special type of $R$-module, thus is actually a tuple $I = (|I|,+,0,\cdot)$ consisting of an underlying set $|I|$, an addition operation $+ : |I| \times |I| \to |I|$, an element $0 \in |I|$, as well as a scalar multiplication $\cdot : |R| \times |I| \to |I|$. Actually, an ideal also comes equipped with a monomorphism $I \to R$, and we also should add the subtraction operator $-$ to the data (even though they are uniquely determined), but let's ignore this for now, also because the notion of a finitely generated ideal does not use the monomorphism $I \to R$ at all.

A ring, in contrast, is a tuple $R = (|R|,+,0,1,\cdot)$ consisting of the data you expect. So the "ideal $R \subseteq R$" is actually $(|R|,+,0,\cdot)$. You forget something, namely the unit. The ideal we are talking about is not equal to $R$.

The notion of a "finitely generated ring" expects as an input a ring, and an ideal does not have all the structure that a ring needs.

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    I have worked a lot in categorical algebra where these distinctions are necessary and also quite common in the literature, in particular when you replace $\mathbf{Set}$ by a well-behaved category. I don't agree with your definition, since you cannot do any algebra with a set (you said subset), but even with your definition the answer is easy: the notion of a "finitely generated ring" expects a ring as an input, and a set (and hence an ideal with your definition) has no ring structure, so that's that. – Martin Brandenburg Dec 30 '23 at 02:47
  • @MartinBrandenburg I'm a bit confused by your answer 2: your definition of an ideal looks like the definition of a module -- a set, with the structure of an abelian group and an action of $R$. Do you mean for an ideal to be a monomorphism $\iota\colon I \to R$ of $R$-modules? Even this seems slightly more general than the "standard" notion of an ideal, because $\iota$ does not have to be the inclusion map (i.e. the terminal object in the arrow category of monomorphisms from $I$ to $R$). Is there a reason to require this relaxation of the normal notion? – krm2233 Jan 11 '24 at 20:13