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This topic I have 2 problem.

Show that if $R$ be Noether ring then $R\neq R[x]$.

And give me a counterexample that above statement wrong if $R$ is not Noether.

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closed as off-topic by Thursday, Ivo Terek, Mark Fantini, Jonas Meyer, RghtHndSd Aug 26 at 2:02

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You clearly did not learn from the comments to your previous questions. Please add your own work and attempts, and don't use the imperative "show". –  akkkk Nov 27 '12 at 14:17
    
The problem is badly stated. If $x \in R$, then of course as rings $R = R[x]$ whether or not $R$ is Noetherian. Presumably you should state that $x$ is an indeterminate (transcendental over $R$), and the question is whether is such case it is possible for $R$ and $R[x]$ to be isomorphic, as they will not be equal (by construction). –  hardmath Nov 27 '12 at 14:43

2 Answers 2

(1) Using the other question you asked: For any $R$, we have the map $\phi\colon R[X] \to R$, $\sum_{k=0}^n a_kX^k \to a_0$, which is always an epimorphism, but never an isomorphism. If you $\psi\colon R \to R[X]$ is an isomorphism, $\phi\psi \colon R \to R$ is an epimorphism, which is no isomorphism. As for $R$ Noetherian such a morphism cannot exist, for Noetherian $R$ we have $R \not\cong R[X]$.

(2) Let for example $k$ a field and $R = k[X_n \mid n \in \mathbb N]$ a polynomial ring in countable many inderterminates. Then $R[Y] \cong R$ via $X_n \mapsto X_{n+1}$, $Y \mapsto X_0$.

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" allways an epimorphism, but never an homomorphism." I believe you mean never an isomorphism. –  JSchlather Nov 27 '12 at 15:27
    
@JacobSchlather Thx. Will edit. –  martini Nov 27 '12 at 15:28

For a counterexmple take $R$ to be the zero ring, then $R[X]$ is a zero ring too. As a bonus, I think this is a Noetherian ring, so it is a counterexample to the statement you do want to prove as well.

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