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I am self studying an online course in algebraic geometry and I am struck on following problem. Unfortunately, I am not very good in solving problems related to noetherian and artinian modules. So, I am asking for help here.

Let A be a ring and let a be a non-zero ideal in A.

(a) If A is a noetherian ring, then every surjective ring endomorphism of A is an automorphism.

Thoughts: Let $f : A \to A$ be an endomorphism. So, I have to prove it 1-1 . let f(a) =f(a') , I am not sure how should I use the property that ring is noetherian to prove that a=a'.Can you please give some hint?

(b) If A is a noetherian ring, then A and A/a are not isomorphic rings.

I assumed that Let they be isomorphic and f be such an isomorphism. But, I am not able to get any Intuition on which property of noetherian rings to use to get a contradiction.

(c) If A is a finite type commutative algebra over the ring R, then every surjective R-algebra endomorphism $\phi$ of A is an automorphism.

Thoughts: Assuming $\phi(x) =0$ I have to prove that x =0.Now, $\phi : A \to A$ is surjective map which is both a ring homo. and R-module homo. Also, A is finite type A-algebra which means that there exists a surjective map $ f : R[x_1,..., x_n] \to A$ . But still I am not able to proceed on what to prove?

So, can you please help me with this question?

I shall be really thankful.

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    (a): see duplicate. (b): The composite of the quotient $A\to A/a$ and the isomorphism $A/a\cong A$ gives a surjective endormorphism of $A$. Apply (a), conclude. (c): Apply the Hilbert basis theorem to see that $A$ is noetherian as well, apply (a). Next, please stop abbreviating "homomorphism" as "homo." – KReiser Dec 08 '21 at 10:42
  • @KReiser Ok will not abbreviate it. –  Dec 08 '21 at 10:43
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    @Jimmy The common abbreviation of "homomorphism" is "hom". – Bill Dubuque Dec 08 '21 at 15:23
  • @KReiser I have understood (a). Thanks to link. –  Jan 19 '22 at 12:59
  • @KReiser For (b) Let there exists an isomorphism i from A to A/a , then it is surjective ring hom from A/a to A . (a) implies that it is an aut from A/a to A/a and similarly an Aut can be defined from A to A. But what is the contracdiction? –  Jan 19 '22 at 13:02
  • @KReiser (c) is done. –  Jan 19 '22 at 13:17
  • The composite of the natural projection $A\to A/a$ and the isomorphism $A/a\cong A$ is a surjective endomorphism of $A$, so it must be an isomorphism, but if $a\neq 0$ it has a nontrivial kernel which is nonsense. – KReiser Jan 20 '22 at 05:55
  • @KReiser But the composition is a map from A to A . So, how a being non zero is a contradiction? –  Jun 08 '22 at 07:00
  • @Avenger Can an isomorphism of rings send an element which is not zero to zero? (Separately, waiting six months to ask a question like this can't be the most effective learning strategy out there and it feels pretty weird from the answering side too.) – KReiser Jun 08 '22 at 13:54

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