Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thursday, Ivo Terek, Mark Fantini, Jonas Meyer, RghtHndSd
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $X_n \mapsto X_{n+1}$, $Y \mapsto X_0$.

share|improve this answer
" 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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.