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Field to ring homomorphism



Know that here we have

f(k1+k2)=f(k1)+f(k2)f(k_1+k_2)=f(k_1)+f(k_2)
f(k1⋅k2)=f(k2)⋅f(k2)f(k_1 \cdot k_2)=f(k_2) \cdot f(k_2)

for k1,k2∈Kk_1, k_2 \in \mathbb{K}

also I know that ff is injective iff kerf={eR}\mathrm{ker} f = \{ e_R \} where eRe_R is the identity element of R.

So f(k)=f(k⋅1)=f(k)⋅f(1)f(k)=f(k \cdot 1)=f(k)\cdot f(1) but for this to be 0 then we need f(1)=0f(1)=0 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.
Reply 1
Avatar for Pangol
Pangol
15
Original post by RDKGames


Know that here we have

f(k1+k2)=f(k1)+f(k2)f(k_1+k_2)=f(k_1)+f(k_2)
f(k1⋅k2)=f(k2)⋅f(k2)f(k_1 \cdot k_2)=f(k_2) \cdot f(k_2)

for k1,k2∈Kk_1, k_2 \in \mathbb{K}

also I know that ff is injective iff kerf={eR}\mathrm{ker} f = \{ e_R \} where eRe_R is the identity element of R.

So f(k)=f(k⋅1)=f(k)⋅f(1)f(k)=f(k \cdot 1)=f(k)\cdot f(1) but for this to be 0 then we need f(1)=0f(1)=0 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.


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.

As a get-out clause, I haven't thought about this sort of thing for years...
Original post by RDKGames
So f(k)=f(k⋅1)=f(k)⋅f(1)f(k)=f(k \cdot 1)=f(k)\cdot f(1) but for this to be 0 then we need f(1)=0f(1)=0 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.?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 1∈K1 \in K is getting mapped to 0∈R0 \in R, 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?
For a ring homomorphism, the kernel is the set of elements mapped to 0 (identity under addition). This is essentially "by definition".
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.Depends on whether your rings contain 1 (by definition) or not.
Reply 4
Avatar for Pangol
Pangol
15
Original post by DFranklin
Depends on whether your rings contain 1 (by definition) or not.


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?
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?
Not really There are effectively two slightly different definitions for a "ring homomorphism" (*):

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.
(edited 6 years ago)
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 1∈K1 \in K is getting mapped to 0∈R0 \in R, 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".


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? Does that mean the function puts all other elements from K to 0 as well - hence the constant function? Then if that's the case f must otherwise be injective as f(0) goes to 0 and f(1) goes to something else, and the same for all other elements of K?

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...
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.
Why do you guess this? (Totally confused myself now).

Does that mean that both additive and mult. identities in K get mapped to the additive identity in R?
No.

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.
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).

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...
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...
Original post by DFranklin
Why do you guess this? (Totally confused myself now).


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; but whatever, I'll stick to the ones without mult. identity.

Spoiler



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).


OK, so say that for some x≠yx \neq y we have f(x)=f(y)f(x)=f(y) which leads us to saying that 0=f(x)−f(y)=f(x−y)0=f(x)-f(y)=f(x-y) where a non-zero element in K is mapped to the additive identity in R. Is this a contradiction since a non-zero element is mapped to an identity?

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...


The existence of multiplicative inverses. Not sure how this fits in though.
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.
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).

OK, so say that for some x≠yx \neq y we have f(x)=f(y)f(x)=f(y) which leads us to saying that 0=f(x)−f(y)=f(x−y)0=f(x)-f(y)=f(x-y) where a non-zero element in K is mapped to the additive identity in R. Is this a contradiction since a non-zero element is mapped to an identity?
It's not a contradiction. It can happen.

What you want to show is that if we can find x≠yx \neq y s.t. f(x)=f(y)f(x)=f(y) 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.
Well, we did just discuss the existence of a particular non-zero element...
(edited 6 years ago)
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 x≠yx \neq y s.t. f(x)=f(y)f(x)=f(y) 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 non-zero element...


So since x−y≠0x-y \neq 0 there is an inverse 1x−y\frac{1}{x-y}, and so if we take 0=f(x−y)0=f(x-y) and mult both sides by f(1x−y)f(\frac{1}{x-y}) we get 0=f(1)0=f(1). How does this show that f(z)=0f(z)=0 though?

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.
Original post by RDKGames
So since x−y≠0x-y \neq 0 there is an inverse 1x−y\frac{1}{x-y}, and so if we take 0=f(x−y)0=f(x-y) and mult both sides by f(1x−y)f(\frac{1}{x-y}) we get 0=f(1)0=f(1). How does this show that f(z)=0f(z)=0 though?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?
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?


OH f(1)⋅f(k)=f(k)=0f(1)\cdot f(k) = f(k) = 0 so if it's not injective then it must be a function equal to 0 for all k in K.

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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