Is there an injective ring homomorphism?

In summary, Homework Equations: -A homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}## is not necessarily unique, and may be composed of other homomorphisms. -To find an injective homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}, one must use the Chinese remainder theorem and trial-and-error.
  • #1
Mr Davis 97
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Homework Statement


From ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##

Homework Equations

The Attempt at a Solution


I know how to do this if we assumed that the rings had to be unital. In that case, there can be no non-trivial homomorphism. However, in my book rings don't need unity, and so a homomorphism is defined only such that the operations are preserved. In this case, how do I show that there is no injective homomorphism?
 
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  • #2
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]?
 
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  • #3
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]?
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}##
 
  • #4
Riddled with mistakes
 
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  • #5
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.
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?

For example, wouldn't ##\varphi## such that ##\varphi (0) = 0##, ##\varphi (1) = 10##, and ##\varphi (2) = 5## also be an injective homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##?
 
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  • #6
Riddled with mistakes
 
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  • #7
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
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##?
 
  • #8
My argument is wrong. [itex]f[/itex] need not be 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.
 
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  • #9
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.
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?
 
  • #10
Riddled with mistakes
 
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  • #11
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?

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.
 
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  • #12
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.
that is not even well-defined :/
 
  • #13
nuuskur said:
that is not even well-defined :/

Sorry, typo. I corrected it.
 
  • #14
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.
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##
 
  • #15
Nono, Dick is right.
##f(1\cdot 1) = f(1)f(1) ## must hold, regardless. But it doesn't.
##5 = f(1) = f(1\cdot 1) \neq f(1)f(1) = 10 ##

What an oversight. It turns out you cannot define ##f## this way. The earlier work is not all for naught, you would still have to determine the only possible 3 element subring to be the image of your monomorphism.
 
  • #16
So does this just mean that there is no homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##? Very confused
 
  • #17
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?
 
  • #18
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?
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.

So will neither of them work for the reason that Dick gave?
 
  • #19
I am so sorry Mr Davis 97, it's such an obvious mistake, too :/

Ok, back to business. The method I described earlier in ##Z_3\times Z_5 ## yields a monomorphism and there does, indeed, exist exactly one such monomorphism. You can verify that the ##g ## in my above post is a monomorphism.

you can verify it by checking all combinations, but that can be counter-productive when you are dealing with more complicated mappings. Let us try to do it algebraically instead. Note that ##f(x) = 10x ## (this you can verify)
Let ##x,y\in Z_3 ## Then

##f(x+y) = 10(x+y) = 10x + 10y = f(x)+f(y) ## and ##f(xy) = 10xy = 10\cdot 10 xy = 10x \cdot 10y = f(x)f(y) ##
where I am using the fact that ##10=10\cdot 10 ## in ##Z_{15} ##.

By the same method you can also see why Dick's answer is right. If you defined ##f(0)=0, f(1)=5, f(2)=10 ## i.e ##f(x) = 5x ## then ##f(xy)=5xy\neq 5x 5y ##.
 
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  • #20
You can simply analyze ##f(1)##. Since ##1\cdot 1= 1## you get ##f(1)\cdot (f(1)-1)=0## and ##f(1) \in \{0,3x,5y\}##.
Now rule out every of these cases.
 
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FAQ: Is there an injective ring homomorphism?

1. What is an injective ring homomorphism?

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.

2. How is an injective ring homomorphism different from a regular ring homomorphism?

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.

3. Can there be more than one injective ring homomorphism between two rings?

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.

4. How do you prove the existence of an injective ring homomorphism?

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.

5. What are some examples of injective ring homomorphisms?

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.

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