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Related problems:
1. A slick proof that a field which is finitely generated as a ring is finite
2. An ideal which is not finitely generated

Sorry for this dumb problem; since I am not a student in mathematics.

My idea is:

since $1\in \mathbf{R}$, so for any $a\in \mathbf{R}$, $a = 1\cdot a$.

So any real number in $\mathbf{R}$ can be finitely generated.

I have no idea if this is true. Can anyone guide me about this by simpler but detailed explanation?

Just for modification:

This problem comes from the following:

A field is Noetherian

What does the $\mathbf{R}$ be finitely generated as is also a part I am confused before; therefore, I hope to understand this throught this question.

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sleeve chen
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    What do you mean by "finitely generated"? As a ring? As a field? In some other way? – Eric Wofsey Dec 21 '16 at 04:12
  • @EricWofsey That is one problem I am confused. http://math.stackexchange.com/questions/1079726/a-field-is-noetherian/1079737?noredirect=1#comment4243238_1079737 – sleeve chen Dec 21 '16 at 04:47
  • @EricWofsey I ask the same problem at the link above. The author told me the answer; however, I am confused of what he said, so just ask this. – sleeve chen Dec 21 '16 at 04:49
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    The term "finitely generated" does not have any meaning on its own. It only has a meaning if you specify what kind of algebraic structure you're talking about. – Eric Wofsey Dec 21 '16 at 04:50
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    @sleevechen as for the problem you quote in your first comment, note that ideals of $R$ are exactly the sub-$R$ modules of the ring, $R$, so that topic is also talking about generation as a module, not as a ring. – Adam Hughes Dec 21 '16 at 05:00

2 Answers2

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Your idea shows that $\Bbb R$ is finitely generated as an $\Bbb R$-module, not as a ring. If $x_1,\ldots, x_n$ are ring generators you are allowed to multiply and add them together-i.e. do the ring operations--which means that $\langle x_1,\ldots, x_n\rangle =\Bbb Z[x_1,\ldots, x_n]$. You cannot just multiply by things in $\Bbb R$ outside of the $x_i$ themselves because multiplication by a scalar is a module operation, not a ring operation. But then since this is countable, it is impossible that there is a surjective ring homomorphism from it onto the uncountable $\Bbb R$.

Adam Hughes
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HINT: The sets $\mathbb C, \mathbb R$, and $\mathbb Q$ are NOT finitely generated. The set of complex and real numbers are both uncountable, and any finitely generated group must be countable (it’s a Foundations of Mathematics style proof). Hope it helps.