Know that here we have
for
also I know that is injective iff where is the identity element of R.
So but for this to be 0 then we need but why is this the case? Why is the multiplicative identity in K getting mapped to the additive identity of R? Why not mult. id. to mult. id.?
How do I know what the identity element of R is since it has 2 operations; addition and multiplication?
Confused on the Q overall but those are my thoughts so far.
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RDKGames
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 16102017 15:31

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 16102017 16:03
(Original post by RDKGames)
Know that here we have
for
also I know that is injective iff where is the identity element of R.
So but for this to be 0 then we need but why is this the case? Why is the multiplicative identity in K getting mapped to the additive identity of R? Why not mult. id. to mult. id.?
How do I know what the identity element of R is since it has 2 operations; addition and multiplication?
Confused on the Q overall but those are my thoughts so far.
But if this is the case, then injectivity follows easily using the fact that f(1) = 1 and f(k1.k2) = f(k1)f(k2), and I can't see how we've used the fact that K is a field  this seems to work just as well if it is a ring.
As a getout clause, I haven't thought about this sort of thing for years... 
DFranklin
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 16102017 16:04
(Original post by RDKGames)
So but for this to be 0 then we need but why is this the case? Why is the multiplicative identity in K getting mapped to the additive identity of R? Why not mult. id. to mult. id.?
In general, it *isn't* going to be true that is getting mapped to , but obviously if f(k) = 0 for all k, then it is going to be true. But I don't know if you're assuming that, trying to prove it, or whatever...
I think the obvious thing to ask yourself here is: "what's special about a field". Unless I'm being stupid, once you answer this, proving the result you want is pretty (very) straightforward.
How do I know what the identity element of R is since it has 2 operations; addition and multiplication? 
DFranklin
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 16102017 16:05
(Original post by Pangol)
This does strike me as a bit odd, because I was under the impression that ring homomorphisms send multiplicative identities to multiplicative identities by definition. I think that the f(k) = 0 for all k option is saying that either the homomorphism is trivial (the image is just the set {0}, where it makes sense for 0 to be both the additive and multiplicative identity), or it is injective.
But if this is the case, then injectivity follows easily using the fact that f(1) = 1 and f(k1.k2) = f(k1)f(k2), and I can't see how we've used the fact that K is a field  this seems to work just as well if it is a ring. 
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 16102017 16:10
(Original post by DFranklin)
Depends on whether your rings contain 1 (by definition) or not. 
DFranklin
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 16102017 16:13
(Original post by Pangol)
So would that be the key  we need the additive and multiplicative identities of K to be distinct, and this will be guaranteed if it is a field?
if you define a ring to have 1, then the homomorphism should preserve 1.
If you don't define a ring to have a 1, then it need not preserve 1.
In this case, I think it's clear it's not assumed rings have a 1 (otherwise the case where everything gets mapped to 0 isn't possible).
Edit: (*) I would hesitate to say what I've written is universally accepted  as ever, the notes for the course being taken should be taken as definitive.Last edited by DFranklin; 16102017 at 16:14. 
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 16102017 18:10
(Original post by DFranklin)
I feel this logic is getting rather convoluted. You're asking why things are true, but this seems to be based on some assumptions (that the product is in the kernel, that f(k) isn't 0, etc) that you aren't explicitly making.
In general, it *isn't* going to be true that is getting mapped to , but obviously if f(k) = 0 for all k, then it is going to be true. But I don't know if you're assuming that, trying to prove it, or whatever...
I think the obvious thing to ask yourself here is: "what's special about a field". Unless I'm being stupid, once you answer this, proving the result you want is pretty (very) straightforward.
For a ring homomorphism, the kernel is the set of elements mapped to 0 (identity under addition). This is essentially "by definition".
Again, not sure where to start with the question so my mind's flying all over the definitions and such but that's what I think the question wants me to show in a nutshell, unless I'm wrong.
The only thing I see that is special about a field is what it has over a ring; so multiplication is commutative, there is a multiplicative identity 1, and every nonzero element in K has an inverse in K. I have already spent over 10 mins looking at these and nothing has clicked as to how they tie in... 
DFranklin
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 16102017 18:22
(Original post by RDKGames)
The def for a ring I got doesn't mention 1 so I guess the one in reference has no mult. identity.
Does that mean that both additive and mult. identities in K get mapped to the additive identity in R?
Again, not sure where to start with the question so my mind's flying all over the definitions and such but that's what I think the question wants me to show in a nutshell, unless I'm wrong.
The only thing I see that is special about a field is what it has over a ring; so multiplication is commutative, there is a multiplicative identity 1, and every nonzero element in K has an inverse in K. I have already spent over 10 mins looking at these and nothing has clicked as to how they tie in... 
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 16102017 18:48
(Original post by DFranklin)
Why do you guess this? (Totally confused myself now).
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(Original post by DFranklin)
I'm not even sure what you're trying to show, but that said, I'm pretty sure you're wrong. I would say that worrying about the kernel, or what identities map to is not the way to go. Instead, go right from the basic definition: if f is not injective, then we can find distinct elements x, y with f(x) = f(y).
Well, K could be a multiplicative ring with 1 and two of those things would still be true. There's only one thing that is specifically true about a field... 
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 16102017 19:02
(Original post by RDKGames)
Well my my module hasn't explicitly used 1 in the def of the ring, but I've seen online that there are rings with element 1 under the name of 'rings with identity' so I wasn't sure whether to take the ring in question as one with the element 1 or not.
What you want to show is that if we can find s.t. then in fact f(z) = 0 for every z.
What I would say to you at this point is you should stop thinking about "additive identities" and "multiplicative identities". Think of them as "0" and "1". This is maybe a little informal, but I'd say it's how most people think about these things, and as it stands it seems quite clear you're getting confused between the two identities.
The existence of multiplicative inverses. Not sure how this fits in though.Last edited by DFranklin; 16102017 at 19:04. 
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 16102017 19:18
(Original post by DFranklin)
My point is, there's a difference between "rings are not defined to have an identity", and "rings are defined not to have an identity". (In the first case, rings may or may not have an identity).It's not a contradiction. It can happen.
What you want to show is that if we can find s.t. then in fact f(z) = 0 for every z.
What I would say to you at this point is you should stop thinking about "additive identities" and "multiplicative identities". Think of them as "0" and "1". This is maybe a little informal, but I'd say it's how most people think about these things, and as it stands it seems quite clear you're getting confused between the two identities.
Well, we did just discuss the existence of a particular nonzero element...
I wasn't confused which one is which when it came to the identities, I was just confusing about the mappings from K to R of them. 
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 16102017 19:20
(Original post by RDKGames)
So since there is an inverse , and so if we take and mult both sides by we get . How does this show that though? 
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 16102017 19:29
(Original post by DFranklin)
Well, if f(1) = 0, and we know that f(a)f(b) = f(ab) for all a, b, what does that tell us about f(b) for arbitrary b?
I've looked at this proof earlier online but didn't quite understand where the step with f(z)=0 came from.
Thanks.
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