Wouldn't this be since the (nontrivial) quotient group isn't even a subset of Z?
In Dummit and Foote Ch. 10 Introduction to Rings on page 228 we read:
"2Z is a subring of Z, as is nZ for any integer n. The ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n 2."
Can anyone help me explictly prove that the ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n >= 2.
Peter
Hmmm ... yes, its a thought, definitely
But then the example of D&F is a little peculiar!
I was thinking that they were identifying with a, but of course if you do this you appear to have a subring (i think)
I tried this with Z/3z and it appeared to be both a subgroup and a subring of Z - so maybe you are correct.
Peter
Thanks.
To clarify, are you essentially saying that because there are n elements in Z/nZ and an infinite number in Z there can be no injective mapping? Or is there more to it?
[I guess I am finding it hard to figure out why "you need to show that there is no (injective or even non-zero) group homomorphism from Z/nZ to Z" (post by NonCommAlg - thanks) leads to the conclusion that "the ring Z/nZ is not a subring (or a subgroup) of Z"]
I started looking at the idea of using a homomorphism to establish what we need and was bothered by the following:
In Dummit and Foote section 7.3 we find:
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"Proposition 5: Let R and S be rings and let : R S be a homomorphism
(1) The image of is a subring of S
(2) the kernel of is a subring of S"
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Now, consider the rings Z and Z/2Z
The map : Z Z/2Z defined by : (even integer) ---> and : (odd integer) ---> is a homomorphism.
Thus its image (which is Z/2Z) is a subring
What have I got wrong? This does not square with D&Fs statement that "2Z is a subring of Z, as is nZ for any integer n. The ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n 2."
Can you please help clarify things.
Peter