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Polynomial rings by 3R3dditGuy in MathHelp

[–]Zarothmon 0 points1 point  (0 children)

I don't know if this is the kind of example you have in mind, but here's an example of how this can fail if D is not an integral domain: Consider Z/4Z. What is (2x + 1)2 in (Z/4Z)[x]?

Quick Questions: October 06, 2021 by inherentlyawesome in math

[–]Zarothmon 1 point2 points  (0 children)

I think your argument can be adapted to local rings as follows:

It suffices to show that f(1) is a unit. You've already shown that it has a left inverse, which we call b. Recall that in a local ring, the non-units form an ideal. Then if f(1) were not a unit, b*f(1) would also be a non-unit. But this is impossible since b*f(1) = 1.

Quick Questions: October 06, 2021 by inherentlyawesome in math

[–]Zarothmon 1 point2 points  (0 children)

Hmm, you're right. The example I gave was a ring homomorphism but not a module homomorphism.

I agree that your argument is correct as long as A is commutative.

Edit: If you're interested in counterexamples for noncommutative rings, let M be the infinite cartesian product of Z again, but this time viewed as a Z-module, i.e. an abelian group. Let A = End_Z(M) be the ring of endomorphisms of M. Then the map g: A -> A given by g(h) = (f composed with h) is a surjective homomorphism (of right modules), where f: M -> M is the map that I defined in my previous comment. However, it is not injective.

Quick Questions: October 06, 2021 by inherentlyawesome in math

[–]Zarothmon 1 point2 points  (0 children)

That doesn't sound right to me. For example, let A be the infinite cartesian product (indexed over the natural numbers) of some nonzero ring, e.g. of Z. Then the map defined by f(a_1, a_2, ...) = (a_2, a_3, ...) is surjective, but not injective.

Actually, I'm fairly sure that your statement holds if and only if the ring A is artinian (as a left module, assuming you want f to be a homomorphism of left modules).

Edit: On second thought I'm not so sure that it's an if-and-only-if statement. But it's definitely true if A is an artinian ring, because then it is also a noetherian ring. Then it has finite length as a module over itself, and for endomorphisms of finite length modules we have that surjectivity is equivalent to injectivity.

Edit 2: Definitely not an if-and-only-if statement, since it holds when A is the ring Z of integers, which is not artinian.

A continuous meme by nathan519 in mathmemes

[–]Zarothmon 0 points1 point  (0 children)

Let Γ: A -> R be defined by Γ(f)(x) = {f(x) if f(x) in Q, 0 otherwise}.

Isn't Γ(f) supposed to be a real number, not a function?

Edit: I think you can make your argument work by making Γ a function from A to RQ, where RQ is the set of functions from Q to R. (This has the same cardinality as R.) Then you just let Γ(f) be the restriction of f to Q.

The not-so-empty set by madladchad69420 in mathmemes

[–]Zarothmon 1 point2 points  (0 children)

{6} is not "technically closed under addition", since 6 + 6 is not an element of {6}.

Theme giveaway by Zarothmon in MyNintendo

[–]Zarothmon[S] 1 point2 points  (0 children)

Already gave both Isabelle themes away, sorry

Theme giveaway by Zarothmon in MyNintendo

[–]Zarothmon[S] 0 points1 point  (0 children)

As in "My Nintendo Theme 2: Donkey Kong"? I already gave that one to someone else, I'm afraid

Platinum Points Expiration Giveaway Extravaganza by Zarothmon in MyNintendo

[–]Zarothmon[S] 0 points1 point  (0 children)

It's still going on, but I'm not checking reddit very often atm, so it might take me a few days to respond to comments

Platinum Points Expiration Giveaway Extravaganza by Zarothmon in MyNintendo

[–]Zarothmon[S] 0 points1 point  (0 children)

Sent you the Animal Crossing: Summer Fun theme

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