I think that ##\mu## such that ##\mu (0) = 0##, ##\mu (1) = 5##, and ##\mu (2) = 10## is an injective homomorphism. However, I found this by trial-and-error, and am not completely sure of a systematic way to find the possible homomorphisms from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##nuuskur said:An image of a homomorphism is always a sub-structure. In this case, if [itex]f:\mathbb{Z}_3\to\mathbb{Z}_{15}[/itex] was a ring homomorphism, then [itex]f(\mathbb{Z}_3)[/itex] is a sub-ring. I can certainly think of an injective ring homomorphism [itex]\mathbb{Z}_3\to\mathbb{Z}_{15}[/itex], though. Would you believe if I wrote [itex]\mathbb{Z}_{15}\cong \mathbb{Z}_3\times\mathbb{Z}_5[/itex]?
So basically you noted that ##\mathbb{Z}_{15}\cong \mathbb{Z}_3\times \mathbb{Z}_5##, and used this to your advantage, because ##f(k) = (k,0)## is an obvious injective homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_3\times \mathbb{Z}_5##, and then found the corresponding elements. However, while ##f(k) = (k,0)## is an obvious homomorphism, how do you know that there aren't others?nuuskur said:Verify it (yes, though, your proposition holds)
As I said [itex]Z_{15}\cong Z_3\times Z_5[/itex]. We can construct the injective homomorphism very easily
[itex]f(k) = (k,0)[/itex], where [itex]k=0,1,2[/itex]. Addition and multiplication in [itex]Z_m\times Z_n[/itex] is defined component-wise: [itex](a,b)+(c,d) =(a+c,b+d)[/itex] and [itex](a,b)(c,d) = (ac,bd)[/itex]. Since [itex]\mbox{gcd}(3,5)=1[/itex] the Chinese remainder theorem will tell you what the corresponding elements of [itex](0,0),(1,0),(2,0)[/itex] in [itex]Z_{15}[/itex] are. They are uniquely determined, so your [itex]\mu[/itex] is, in fact, the only possibility.
Could you address the first part of my last question? That is, why is ##f(k) = (k,0)## the only homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_3 \times \mathbb{Z}_5##?nuuskur said:Good catch. The chinese remainder theorem tells you the corresponding elements are uniquely determined, but you are free to combine them however you wish as long as you don't violate the criteria of ring isomorphism.
The image [itex]\{0,5,10\}[/itex] is uniquely determined, but not the mapping itself. Thank you
Well in that case we have two homomorphisms. But how do we know that ##f(k) = (k,0)##, where k = 1,2,3, and the ##g## that you describe are the only two such homomorphisms? How do you knowfor sure that other's can't be constructed?nuuskur said:My argument is wrong. [itex]f[/itex] is not unique. The image of the injective homomorphism (aka monomorphism) is unique, but it doesn't imply the mapping is unique. You proposed [itex]g(0)=0, g(1) = (2,0), g(2) = (1,0)[/itex] and this is also a monomorphism.
Pardon the confusion.
Mr Davis 97 said:Well in that case we have two homomorphisms. But how do we know that ##f(k) = (k,0)##, where k = 1,2,3, and the ##g## that you describe are the only two such homomorphisms? How do you knowfor sure that other's can't be constructed?
that is not even well-defined :/Dick said:Actually, defining the map ##\varphi## such that ##\varphi (0) = 0##, ##\varphi (1) = 5##, ##\varphi (1) = 10##, doesn't give a ring homomorphism. ##\varphi (1*1) = \varphi (1) \varphi (1)## doesn't work.
nuuskur said:that is not even well-defined :/
We are using the definition of the ring such that there isn't necessarily a multiplicative inverse, and hence the definition of homomorphism doesn't include the condition that ##\mu (1_A) = 1_B##, where ##\mu : A \rightarrow B##Dick said:Actually, defining the map ##\varphi## such that ##\varphi (0) = 0##, ##\varphi (1) = 5##, ##\varphi (2) = 10##, doesn't give a ring homomorphism. ##\varphi (1*1) = \varphi (1) \varphi (1)## doesn't work.
I don't see why it's not a subring. It has the zero element, the difference between any two elements is in the set, and the multiplication of any two elements is in the set.nuuskur said:Oh my lord. This is all my fault, I made a huge mistake somewhere in my argumentation. I will attempt to make amends.
1. Image of ring homomorphism is a subring. Since ##Z_3 ## contains three elements, we need a three element subring of ##Z_{15} ##.
2. Because ##(Z_{15},+) ## is cyclic, its three element sub-abelian group must also be cyclic. Since the subgroup contains three elements, it must contain an element of order ##3## modulo ##15##. The only candidate is ##5## and it generates the subgroup ##\{0,5,10\}##. Is it a subring? Yes, it is, verify it.
3. Since a ring homomorphism preserves zero element there are only two candidate monomorphisms
##f(0)=0, f(1)=5, f(2)= 10 ## OR ##g(0)=0, g(1)=10, g(2) =5 ##. Will either of them work?
An injective ring homomorphism is a function between two rings that preserves the ring structure and is one-to-one, meaning that distinct elements in the first ring are mapped to distinct elements in the second ring.
An injective ring homomorphism is a special case of a regular ring homomorphism, where the function is one-to-one. This means that the inverse image of any element in the second ring has at most one element in the first ring, while a regular ring homomorphism can have multiple elements in the inverse image.
Yes, it is possible for there to be multiple injective ring homomorphisms between two rings. However, there can only be one bijective ring homomorphism between two rings.
The existence of an injective ring homomorphism can be proven by constructing a function that preserves the ring structure and is one-to-one. This can be done by defining the function explicitly or by using a known property of the rings.
Some examples of injective ring homomorphisms include the inclusion map from the integers to the rational numbers, the natural embedding from the integers to the Gaussian integers, and the identity map from a ring to itself.