Can real line $\mathbf{R}$ be finitely generated?

Question

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.

Answer 1

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 $\Bbb R$.

Answer 2

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.