How can the kernel of a ring morphism be a subring?

  • #1
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Main Question or Discussion Point

I don't understand this page, https://www.proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Subring, but how can this be a true statement? Wouldn't a ring morphism map the multiplicitive identity to itself? So it wouldn't be in the kernel, so how could the kernel be a subring?

I happened upon this whilst trying to figure out why the kernel of a morphism is an ideal in the pre-image ring or whatever. Anyone enlighten me and alleviate my confusion?
 

Answers and Replies

  • #2
Stephen Tashi
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Wouldn't a ring morphism map the multiplicitive identity to itself? So it wouldn't be in the kernel, so how could the kernel be a subring?
The kernel consists of those elements in the group that are mapped to the identity by the homomorphism. Since the identity is mapped to the identity, it is one of those elements that is mapped to the identiy. Hence the identity is an element of the kernel.
 
  • #3
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Wait, i thought Ker(a) = {a(r) = 0 | for all r in R}. I understand that this is mapping all the elements to the additive identity, but rings also have a multiplicative identity? How does the kernel of a mapping have both 0 and 1 when 0 maps to 0 and 1 maps to 1?
 
  • #4
Stephen Tashi
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Wait, i thought Ker(a) = {a(r) = 0 | for all r in R}. I understand that this is mapping all the elements to the additive identity
You mean "....that this is the set of elements that are mapped to the identity" - but I see what you are getting at. I'd put the question this way:

if we have a homorphism [itex]f [/itex] from ring [itex] R [/itex] to ring [itex] S [/itex] then must [itex] ker(f) [/itex] be a subring of [itex]R [/itex] ? It seems that if mathematicians have defined things in a nice way that it should be.

However, if we define [itex] ker(f) = \{ x: x\in R, f(x)= 0_S \} [/itex] then a multiplicative identity [itex] 1_R [/itex] in [itex] R [/itex] isn't necessarily an element of [itex] ker(f) [/itex]. How shall we fix this? Shall we let [itex] ker(f) [/itex] be a ring by using definition of "ring" that doesn't require it have a multiplicative identity? Shall we change the definition of [itex] ker(f) [/itex] ?

You have to figure out which definition of "ring" and which definition of a "homomorphism between rings" will govern. For example, there is Wikipedia http://en.wikipedia.org/wiki/Ring_homomorphism versus Wolfram http://mathworld.wolfram.com/RingHomomorphism.html. The Wikipedia definition assumes a "ring" has a multiplicative identity. Wolfram uses a definition of a "ring" that doesn't require it to have a multiplicative identity.

What ProofWiki thinks about math is hard to decypher. ProofWiki appears to be an attempt to write minimalist proofs. Such proofs depend on subtle properties of definitions and long chains of theorems. If you try to trace back how ProofWiki defines a "ring", you follow links till you get to a page on semirings that has a warning that it is being "refactored".
 
  • #5
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I'm new to math and algebra in general, but I was under the impression that structures as fundamental to modern algebra as a ring would have a solid definition. Should I not be surprised? Both of the definitions are valid?
 
  • #6
Stephen Tashi
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Both of the definitions are valid?
Mathematics being a cultural endeavor, I'd say both the definitions are "encountered". Within one context, you obviously shouldn't use them simultaneously.

The reliable theorem appears to be that [itex] ker(f) [/itex] of a homomorphism of rings [itex] f:R \rightarrow S [/itex] is an ideal of the ring [itex] R [/itex].
 
  • #7
Erland
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When I learned abstract algebra, I learned that a ring need not have a multiplicative identity. It was a surprise to me that in some texts, such as the Wikipedia, the definition is changed to include a multiplicative identity. Wikipedia calls a structure which I learned is called a ring a rng (for which a mulrtiplicative identity is not assumed). Frankly, I don't understand at all why one should change the old definition of ring.
 

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