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I heard that a field is always Noetherian and here Noetherian means that every ideal is finitely generated. Then, because a field has two ideals, 0 and the field itself, does this mean every field have to be finitely generated? Where I got it wrong?

Matt Samuel
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  • When you say something is finitely generated, you have to specify what you want it to be finitely generated as. All fields are finitely generated as rings. But, for example, $\mathbb Q$ is not finitely generated as an additive group. – Mathmo123 Dec 24 '14 at 12:26
  • @Mathmo123 Not all fields are finitely generated as rings. In fact, only the finite ones are. – Tobias Kildetoft Dec 24 '14 at 13:00
  • @TobiasKildetoft sorry that should say finitely generated as ideals of rings – Mathmo123 Dec 24 '14 at 13:54
  • @Mathmo But that is true for all unital rings. – Tobias Kildetoft Dec 24 '14 at 14:03
  • @TobiasKildetoft that was exactly my point. The OP had misunderstood what it meant to be finitely generated. When we say something is finitely generated, being "finitely generated" depends entirely on the context. I used the ideal example because that is the example in the question, but you're right that it tells us nothing new – Mathmo123 Dec 24 '14 at 14:05

1 Answers1

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"Finitely generated" means finitely generated as an ideal. An ideal $I\subseteq R$ is finitely generated if there exist finitely many elements $x_1,x_2,\ldots,x_n\in I$ such that for every $y\in I$ there exist $r_1,r_2,\ldots,r_n\in R$ such that $$y=r_1x_1+r_2x_2+\cdots+r_nx_n$$ In particular, for any unital ring the set $\{1 \}$ qualifies as a generating set for the ideal consisting of the entire ring.

Matt Samuel
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  • Thanks!! Now I see the difference. –  Dec 24 '14 at 12:39
  • @acconid You're welcome. – Matt Samuel Dec 24 '14 at 12:40
  • One question: Do $x_1,\ldots,x_n$ have to be in the ideal $I$ in your definition of finitely generated ideal? If it is yes, then $1$ is in all ideal consisting the entire ring? – sleeve chen Dec 20 '16 at 23:02
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    @sleeve I guess I didn't mention it, the $x_i$ need to be elements of $I$. $1$ is in $I$ if and only if $I=R$. – Matt Samuel Dec 20 '16 at 23:40
  • One more question, so according to your answer, real number $\mathbf{R}$ is finitely generated since $\forall a\in \mathbf{R}$, $a = a\cdot 1$, i.e. I take $1$ as a generator. So $\mathbf{R}$ is a finite field. However, it is not? I am confused about this. – sleeve chen Dec 21 '16 at 03:44
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    @sleeve The field of real numbers is not finitely generated as a ring, and it is not finitely generated as a field. It is finitely generated as an ideal over itself, which is true of any unital ring. – Matt Samuel Dec 21 '16 at 03:47