Let $A=\{q+r_1X+ \cdots +r_nX^n: q \in \mathbb{Q}, r_i \in \mathbb{R}\}$ be the polynomial ring with rational costant terms. I have to prove that $A$ isn't a noetherian ring. How can I prove it?
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Well, you can either use the definition of Noetherian ring, or some result you know of the form, "If $A$ is Noetherian, then (some statement about $A$)." What results do you know? – Thomas Andrews Oct 06 '13 at 17:01
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1I know that there are three equivalent definitions: ascending chain condition, every ideal is finitely generated and every non empty set of ideals has a maximal element. – ArthurStuart Oct 06 '13 at 17:08
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You subject asks whether it is Noetherian, but you say you are asked to prove it is not. Which is it - are you asked whether $A$ Noetherian, or asked to prove it is not? – Thomas Andrews Oct 06 '13 at 17:16
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Let $R$ be an infinite set of reals which are linearly independent when the reals are considered a vector space over the rationals (such a set exists because the reals are uncountable and the rationals are countable). Then $\langle rx \, | r \in R \rangle$ will be an ideal that is not finitely generated.
user2566092
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consider the chain of ideals:
$\pi xA \subset \pi xA + \pi^2xA \subset\pi xA +\pi^2 xA +\pi^3 xA \subset ...$
user99126
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Consider the ideal $I=\oplus _{i\geq 1}\mathbb R\cdot X^i\subset A$.
Despite appearances it is not equal to the ideal $(X)=XA$: we only have the inclusion $(X)\subsetneq I$ !
For example $\sqrt 2X\in I\setminus (X)$.
Once you understand this, you will quickly be able to see why $I$ is not finitely generated by considering the sequence $(r_jX)_j$ of polynomials in $I$, where the $r_j\in \mathbb R$ are linearly independent over $\mathbb Q$ .
Georges Elencwajg
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