# Is there an injective ring homomorphism?

1. Mar 22, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
From $\mathbb{Z}_3$ to $\mathbb{Z}_{15}$

2. Relevant equations

3. 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?

Last edited: Mar 23, 2017
2. Mar 23, 2017

### 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 $\mathbb{Z}_{15}\cong \mathbb{Z}_3\times\mathbb{Z}_5$?

Last edited: Mar 23, 2017
3. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

### nuuskur

Riddled with mistakes

Last edited: Mar 23, 2017
5. Mar 23, 2017

### Mr Davis 97

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 $\mathbb{Z}_{15}$?

6. Mar 23, 2017

### nuuskur

Riddled with mistakes

Last edited: Mar 23, 2017
7. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

### 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.

Last edited: Mar 23, 2017
9. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

### nuuskur

Riddled with mistakes

Last edited: Mar 23, 2017
11. Mar 23, 2017

### Dick

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.

12. Mar 23, 2017

### nuuskur

that is not even well-defined :/

13. Mar 23, 2017

### Dick

Sorry, typo. I corrected it.

14. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

### 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.

16. Mar 23, 2017

### Mr Davis 97

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

17. Mar 23, 2017

### 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?

18. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

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

Last edited: Mar 23, 2017
20. Mar 23, 2017

### Staff: Mentor

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.