# Ring Homomorphisms

• Sep 5th 2011, 02:25 AM
Bernhard
Ring Homomorphisms
Can anyone help me with an effective approach to this problem

Find all possible ring homomorphisms betweeen the indicated rings

(a) $\phi$ : $Z_6$ $\rightarrow$ $Z_{10}$

(b) Q( $\surd$ 2) $\rightarrow$ Q( $\surd$ 3)

Would be most grateful of help.

Peter
• Sep 5th 2011, 02:59 AM
Swlabr
Re: Ring Homomorphisms
Quote:

Originally Posted by Bernhard
Can anyone help me with an effective approach to this problem

Find all possible ring homomorphisms betweeen the indicated rings

(a) $\phi$ : $Z_6$ $\rightarrow$ $Z_{10}$

(b) Q( $\surd$ 2) $\rightarrow$ Q( $\surd$ 3)

Would be most grateful of help.

Peter

If $\phi: \mathbb{Z}_6\rightarrow \mathbb{Z}_{10}$ then note that there are no elements of order 3 in $\mathbb{Z}_{10}$. So, if $\mathbb{Z}_{6}$ is to map anywhere non-trivially it must map to a copy of $\mathbb{Z}_2$. So, do this and see where it takes you.

There are no non-trivial ones between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. This is a classic question - basically, show that 0 must map to 0, and 1 to 1, but 1+1=2, which has a square root in the pre-image, and 1+1+1=3 which has a square root in the image. This is a contradiction (can you see why?). I will leave it to you to formalise it all.
• Sep 5th 2011, 11:16 AM
Drexel28
Re: Ring Homomorphisms
Quote:

Originally Posted by Swlabr
If $\phi: \mathbb{Z}_6\rightarrow \mathbb{Z}_{10}$ then note that there are no elements of order 3 in $\mathbb{Z}_{10}$. So, if $\mathbb{Z}_{6}$ is to map anywhere non-trivially it must map to a copy of $\mathbb{Z}_2$. So, do this and see where it takes you.

Or, perhaps easier is to note that if $f:R\to S$ is a unital homomorphism then $\text{char}(S)\mid\text{char}(R)$. More down to earth, if $\phi:\mathbb{Z}_{6}\to\mathbb{Z}_{10}$ is a unital morphism then it maps the generator $1$ to the generator $1$ and so, in particular, $\phi$ is surjective...but I'm sure it's easy to see this isn't possible.
• Sep 6th 2011, 01:39 AM
Swlabr
Re: Ring Homomorphisms
Quote:

Originally Posted by Drexel28
Or, perhaps easier is to note that if $f:R\to S$ is a unital homomorphism then $\text{char}(S)\mid\text{char}(R)$. More down to earth, if $\phi:\mathbb{Z}_{6}\to\mathbb{Z}_{10}$ is a unital morphism then it maps the generator $1$ to the generator $1$ and so, in particular, $\phi$ is surjective...but I'm sure it's easy to see this isn't possible.

Don't both fields have characteristic zero? I forget - fields are a distant memory...
• Sep 6th 2011, 12:47 PM
Drexel28
Re: Ring Homomorphisms
Quote:

Originally Posted by Swlabr
Don't both fields have characteristic zero? I forget - fields are a distant memory...

Yeah, it's hard to remember details sometime. $\mathbb{Z}_6$ and $\mathbb{Z}_{10}$ are not fields, the only quotient rings of $\mathbb{Z}$ which are fields are $\mathbb{Z}_p$ where $p$ is prime. And, no, the characteristic of a unital ring is it's order in the additive group. So, for example $\text{char}(\mathbb{Z}_6)=|1|_{\mathbb{Z}_6}=6$ and $\text{char}(\mathbb{Z}_{10})=|1|_{\mathbb{Z}_{10}} =10$.
• Sep 7th 2011, 02:05 AM
Swlabr
Re: Ring Homomorphisms
Quote:

Originally Posted by Drexel28
Yeah, it's hard to remember details sometime. $\mathbb{Z}_6$ and $\mathbb{Z}_{10}$ are not fields, the only quotient rings of $\mathbb{Z}$ which are fields are $\mathbb{Z}_p$ where $p$ is prime. And, no, the characteristic of a unital ring is it's order in the additive group. So, for example $\text{char}(\mathbb{Z}_6)=|1|_{\mathbb{Z}_6}=6$ and $\text{char}(\mathbb{Z}_{10})=|1|_{\mathbb{Z}_{10}} =10$.

Sorry, I miss-interpreted your post. I thought you were talking about the characteristics of $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ (which are zero, and so my memory is working!).
• Sep 7th 2011, 02:33 AM
Bernhard
Re: Ring Homomorphisms
Thanks to those posting help ... but I am still struggling to determine an answer ...

"Find all possible homomorphisms between the indicated groups: $\phi$: $Z_6$ $\rightarrow$ $Z_{10}$"

Swlabr writes: (thanks for help by the way!)

"If $\phi$: $Z_6$ $\rightarrow$ $Z_{10}$, then note that there are no elements of order 3 in $Z_{10}$. So if $Z_6$ is to map anywhere nontrivially it must map to a copy of $Z_2$. So do this and see where it takes you."

However as far as I got was as follows:

I can see that the homomorphism $\phi$ is completely determined by $\phi$( $[1]_{10}$) = $[m]_{10}$

and, since $[1]_6$ has order 6 then we have 0( $[m]_{10}$) | 6

Thus 0( $[m]_{10}$) is 6, 3,2 or 1 (Correct?)

... but ... what exactly are the next few steps?

Would be grateful for some help

Peter
• Sep 7th 2011, 03:50 AM
Bernhard
Re: Ring Homomorphisms
Just a note:

I can see that if you check all elements of $Z_{10}$ that you find that only the element $[5]_{10}}$ has an order that divides 6.

But what is a convenient way to do this if enumeration is difficult?

Peter
• Sep 7th 2011, 03:52 AM
Opalg
Re: Ring Homomorphisms
Quote:

Originally Posted by Bernhard
Thanks to those posting help ... but I am still struggling to determine an answer ...

"Find all possible homomorphisms between the indicated groups: $\phi$: $Z_6$ $\rightarrow$ $Z_{10}$"

Swlabr writes: (thanks for help by the way!)

"If $\phi$: $Z_6$ $\rightarrow$ $Z_{10}$, then note that there are no elements of order 3 in $Z_{10}$. So if $Z_6$ is to map anywhere nontrivially it must map to a copy of $Z_2$. So do this and see where it takes you."

However as far as I got was as follows:

I can see that the homomorphism $\phi$ is completely determined by $\phi$( $[1]_{10}$) = $[m]_{10}$

and, since $[1]_6$ has order 6 then we have 0( $[m]_{10}$) | 6

Thus 0( $[m]_{10}$) is 6, 3,2 or 1 (Correct?)

... but ... what exactly are the next few steps?

The first essential is that you need to be clear about whether you are looking for group homomorphisms or ring homomorphisms. In a previous posting, you asked about group homomorphisms from $\mathbb{Z}_6$ to $\mathbb{Z}_{10}.$ As you correctly say in the above quote, if $\phi: \mathbb{Z}_6\to\mathbb{Z}_{10}$ is a group homomorphism with $\phi ([1]_{6})= [m]_{10},$ then [m] must have order 6, 3, 2 or 1 in the additive group of integers mod 10. But that group has no elements of order 3 or 6. Therefore [m] must have order 1 or 2.

But there is an extra complication if you are looking for a ring homomorphism, because ring homomorphisms have to preserve multiplication (in this case, multiplication mod 10) as well as addition. In fact, a ring homomorphism between unital rings is either identically zero, or it has to take the identity element to an idempotent element. In this case, that means that [m] must be [0], [1] or [5]. But it can't be [1] because that has order 10... .