Is there an injective ring homomorphism?

[Mr Davis 97]

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?

 

[nuuskur]

An image of a homomorphism is always a sub-structure. In this case, if f:\mathbb{Z}_3\to\mathbb{Z}_{15} was a ring homomorphism, then f(\mathbb{Z}_3) is a sub-ring. I can certainly think of an injective ring homomorphism \mathbb{Z}_3\to\mathbb{Z}_{15}, though. Would you believe if I wrote \mathbb{Z}_{15}\cong \mathbb{Z}_3\times\mathbb{Z}_5?

 

[Mr Davis 97]

An image of a homomorphism is always a sub-structure. In this case, if f:\mathbb{Z}_3\to\mathbb{Z}_{15} was a ring homomorphism, then f(\mathbb{Z}_3) is a sub-ring. I can certainly think of an injective ring homomorphism \mathbb{Z}_3\to\mathbb{Z}_{15}, though. Would you believe if I wrote \mathbb{Z}_{15}\cong \mathbb{Z}_3\times\mathbb{Z}_5?

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]

Riddled with mistakes

 

[Mr Davis 97]

Verify it (yes, though, your proposition holds)

As I said Z_{15}\cong Z_3\times Z_5. We can construct the injective homomorphism very easily
f(k) = (k,0), where k=0,1,2. Addition and multiplication in Z_m\times Z_n is defined component-wise: (a,b)+(c,d) =(a+c,b+d) and (a,b)(c,d) = (ac,bd). Since \mbox{gcd}(3,5)=1 the Chinese remainder theorem will tell you what the corresponding elements of (0,0),(1,0),(2,0) in Z_{15} are. They are uniquely determined, so your \mu 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 $\mathbb{Z}_{15}$?

 

[nuuskur]

Riddled with mistakes

 

[Mr Davis 97]

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 \{0,5,10\} 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 $\mathbb{Z}_3 \times \mathbb{Z}_5$?

 

[nuuskur]

My argument is wrong. f 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 g(0)=0, g(1) = (2,0), g(2) = (1,0) and this is also a monomorphism.

Pardon the confusion.

 

[Mr Davis 97]

My argument is wrong. f is not unique. The image of the injective homomorphism (aka monomorphism) is unique, but it doesn't imply the mapping is unique. You proposed g(0)=0, g(1) = (2,0), g(2) = (1,0) 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?

 

[nuuskur]

Riddled with mistakes

 

[Dick]

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.

 

[nuuskur]

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 :/

 

[Dick]

that is not even well-defined :/

Sorry, typo. I corrected it.

 

[Mr Davis 97]

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 $\mu : A \rightarrow B$

 

[nuuskur]

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.

 

[Mr Davis 97]

So does this just mean that there is no homomorphism from $\mathbb{Z}_3$ to $\mathbb{Z}_{15}$? Very confused

 

[nuuskur]

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?

 

[Mr Davis 97]

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?

 

[nuuskur]

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 $f(xy)=5xy\neq 5x 5y$.

 

[fresh_42]

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.