# Ring Homomorphism

• Sep 21st 2010, 02:57 AM
demode
Ring Homomorphism
http://img28.imageshack.us/img28/2453/69935252.gif

(a) What does the question mean by 'determining the kernel'? I know that kernels are ideals and ideals are kernels. Since $\varphi$ is a ring homomorphism from $R \to \mathbb{Z}$, then $ker(\varphi)=\{ r \in R | \varphi (r) = 0 \}$ is an ideal of R.

Now, a subring I is an ideal of R if $rI := \{ ra: a \in I \} \subseteq I$ and $Ir := \{ ar : a \in I \} \subseteq I$ for $\forall r \in R$. (The subring I is closed under multiplication with arbitrary elements of R). Is this all I need to say?!

(b) I have no idea how to do this one...

(c) I think $ker(\varphi)$ is a prime ideal of R, if it is a proper ideal of R such that $a,b \in R$ and $ab \in ker(\varphi)$ imply $a \in ker(\varphi)$ or $b \in ker(\varphi)$. What else do I need to state?
• Sep 21st 2010, 03:30 AM
Swlabr
Quote:

Originally Posted by demode
http://img28.imageshack.us/img28/2453/69935252.gif

(a) What does the question mean by 'determining the kernel'? I know that kernels are ideals and ideals are kernels. Since $\varphi$ is a ring homomorphism from $R \to \mathbb{Z}$, then $ker(\varphi)=\{ r \in R | \varphi (r) = 0 \}$ is an ideal of R.

Now, a subring I is an ideal of R if $rI := \{ ra: a \in I \} \subseteq I$ and $Ir := \{ ar : a \in I \} \subseteq I$ for $\forall r \in R$. (The subring I is closed under multiplication with arbitrary elements of R). Is this all I need to say?!

No. I mean, you've just given the definition of a kernel. You need to apply it to this case. It's just the set of matrices such that a-3b=0. So just stick that in set notation...

Quote:

Originally Posted by demode
(b) I have no idea how to do this one...

Okay, you have a ring homomorphism which maps into the integers. Basically, you are trying to show that the image is isomorphic to the integers (this is easiest, and equivalent). So, how would you approach this problem? You know that the image injects into the integers (because it is a subring). So, what is left to show?...

Quote:

Originally Posted by demode
(c) I think $ker(\varphi)$ is a prime ideal of R, if it is a proper ideal of R such that $a,b \in R$ and $ab \in ker(\varphi)$ imply $a \in ker(\varphi)$ or $b \in ker(\varphi)$. What else do I need to state?

This is a exercise in looking at the image to find things about the pre-image/kernel. So, if $ker(\varphi)$ is a prime ideal, then what does this mean about the integers? If $ker(\varphi)$ is a maximal ideal what does this mean about the integers?
• Sep 21st 2010, 06:08 PM
demode
Quote:

Originally Posted by Swlabr
No. I mean, you've just given the definition of a kernel. You need to apply it to this case. It's just the set of matrices such that a-3b=0. So just stick that in set notation...

$ker(\varphi)= \{ a-3b=0 | a,b \in \mathbb{Z} \}$

Quote:

Okay, you have a ring homomorphism which maps into the integers. Basically, you are trying to show that the image is isomorphic to the integers (this is easiest, and equivalent). So, how would you approach this problem? You know that the image injects into the integers (because it is a subring). So, what is left to show?...
So, I just need to show that $R/ker(\varphi)$ is surjective and operation preserving?

Quote:

This is a exercise in looking at the image to find things about the pre-image/kernel. So, if $ker(\varphi)$ is a prime ideal, then what does this mean about the integers? If $ker(\varphi)$ is a maximal ideal what does this mean about the integers?
What do you mean by that? Could you please explain a little bit more?
• Sep 22nd 2010, 12:14 AM
Swlabr
Quote:

Originally Posted by demode
So, I just need to show that $R/ker(\varphi)$ is surjective and operation preserving?

You have operation preserving - that's just the first Isomorphism theorem. So you just need to prove surjective. Can you think of a neat way of proving that the homomorphism is surjective?

Quote:

Originally Posted by demode
What do you mean by that? Could you please explain a little bit more?

$I$ is a prime ideal in $R$
$\Leftrightarrow$ $ab \in I \Rightarrow a\in I$ or $b \in I$
$\Leftrightarrow$...what does this mean about $R/I$? Hint: zero divisors.
• Sep 23rd 2010, 01:53 AM
demode
Quote:

$I$ is a prime ideal in $R$
$\Leftrightarrow$ $ab \in I \Rightarrow a\in I$ or $b \in I$
$\Leftrightarrow$...what does this mean about $R/I$? Hint: zero divisors.
I tried to follow your hint, but I still need some help to wrap it up:

If $I$ is a prime ideal in $R$ then: $R/I$ is an integral domain if and only if $I$ is a prime ideal. (Also $R/I$ is a field if and only if $I$ is maximal.)

$R/I$ is clearly commutative and it has identity $1+I$. In order to show it's an integral domain I only need to show that there are no zero divisors.

x+I and y+I are nonzero in $R/I$ $\iff$ $x$ and $y$ do not lie in $I$. If I is a prime then x.y dos not lie in I and so $(x+I)(y+I)=x.y+I$ must be non-zero.

Let $x= \begin{pmatrix}a_1 &3b_1 \\3b_1 & a_1 \end{pmatrix}$, $y= \begin{pmatrix}a_2 &3b_2 \\3b_2 & a_2 \end{pmatrix}$

$xy= \begin{pmatrix}a_1a_2+9b_1b_2 &3a_1b_2+3a_2b_1 \\3a_2b_1+3a_1b_2 & 9b_1b_2+a_1a_2 \end{pmatrix}$

$\varphi (xy)= a_1a_2 +9b_1b_2 - 3(a_1b_1+a_2b_1)$

$=(a_1-3b_1)(a_23b_2)=0$

$xy \in ker(\varphi)$

$\implies a_1-3b_1 = 0$ or $a_2-3b_2=0$

So $x \in ker(\varphi)$ or $y \in ker(\varphi)$

What else do I need to do? Am I right so far?

Quote:

Originally Posted by Swlabr
You have operation preserving - that's just the first Isomorphism theorem. So you just need to prove surjective. Can you think of a neat way of proving that the homomorphism is surjective?

I see, but I really can't think of a neat way of doing it... maybe more hints please?
• Sep 23rd 2010, 01:57 AM
Swlabr
Quote:

Originally Posted by demode
I tried to follow your hint, but I'm not sure how to wrap it up:

If $I$ is a prime ideal in $R$ then:

$R/I$ is an integral domain if and only if $I$ is a prime ideal.

(Also $R/I$ is a field if and only if $I$ is maximal.)

$R/I$ is clearly commutative and it has identity $1+I$. In order to show it's an integral domain I only need to show that there are no zero divisors.

x+I and y+I are nonzero in $R/I$ $\iff$ $x$ and $y$ do not lie in $I$. If I is a prime then x.y dos not lie in I and so

$(x+I)(y+1)=x.y+I$

must be non-zero.

Let $x= \begin{pmatrix}a_1 &3b_1 \\3b_1 & a_1 \end{pmatrix}$, $y= \begin{pmatrix}a_2 &3b_2 \\3b_2 & a_2 \end{pmatrix}$

$xy= \begin{pmatrix}a_1a_2+9b_1b_2 &3a_1b_2+3a_2b_1 \\3a_2b_1+3a_1b_2 & 9b_1b_2+a_1a_2 \end{pmatrix}$

$\varphi (xy)= a_1a_2 +9b_1b_2 - 3(a_1b_1+a_2b_1)$

$=(a_1-3b_1)(a_23b_2)=0$

$xy \in ker(\varphi)$

$\implies a_1-3b_1 = 0 or a_2-3b_2=0$

So $x \in ker(\varphi)$ or $y \in ker(\varphi)$

What else should I do?

Use the fact that $R/I \cong \mathbb{Z}$. No working is really needed then.

Quote:

Originally Posted by demode
I see, but I really can't think of a neat way of doing it... maybe some more hints please?

You need to prove that the generators (or here, generator) is mapped to. Can you see why this is sufficient?
• Sep 24th 2010, 05:28 AM
demode
Quote:

Originally Posted by Swlabr
Use the fact that $R/I \cong \mathbb{Z}$. No working is really needed then.

I'm not sure how to fit this in, could you please show me where it can be used? Here's the proof I've written so far:

Suppose I is a prime and suppose R/I is not an integral domain. There are zero divisors in R/I iff there are $(x+I), (y+I) \in (R/I)^*$. So that $(x+I)(y+I) = 0_R =I \in (R/I)^*$.

Now, $(x+I) \in (R/I)^* \iff x+I \neq I \iff x+I \notin I \iff x \notin I$.

$y+I \in (R/I)^* \iff y \notin I$

but $xy+I=I \iff xy \in I$

So, $x \notin I$and $y \notin I$ but $xy \in I$. So we get a contradiction and R/I must be an integral domain and so $I=ker(\varphi)$ is a prime.

Quote:

You need to prove that the generators (or here, generator) is mapped to. Can you see why this is sufficient?
I'm not quiete sure what you mean. But here's what I've tried:

$R/I \cong \mathbb{Z}$, so $(x+I) \mapsto x$. Let $x+I \in R/I$ where

$x+I = \begin{pmatrix}a &3b \\3b & a \end{pmatrix} + I$

Since $a-3b=0 \iff a= 3b$,

$I = \left\{ \begin{pmatrix}3b &3b \\3b & 3b \end{pmatrix} | b \in \mathbb{Z} \right\}$. Therefore

$\varphi \left( \begin{pmatrix}a &3b \\3b & a \end{pmatrix} + I \right) = \varphi \left( \begin{pmatrix}a &3b \\3b & a \end{pmatrix} - \begin{pmatrix}3b &3b \\3b & 3b \end{pmatrix} \right)= \varphi \left( \begin{pmatrix}a-3b &0 \\0 & a-3b \end{pmatrix} + I \right)$

= (a-3b)-(a-3b)=0

Is this also correct?
• Sep 27th 2010, 06:26 AM
Swlabr
Quote:

Originally Posted by demode
I'm not sure how to fit this in, could you please show me where it can be used? Here's the proof I've written so far:

Suppose I is a prime and suppose R/I is not an integral domain. There are zero divisors in R/I iff there are $(x+I), (y+I) \in (R/I)^*$. So that $(x+I)(y+I) = 0_R =I \in (R/I)^*$.

Now, $(x+I) \in (R/I)^* \iff x+I \neq I \iff x+I \notin I \iff x \notin I$.

$y+I \in (R/I)^* \iff y \notin I$

but $xy+I=I \iff xy \in I$

So, $x \notin I$and $y \notin I$ but $xy \in I$. So we get a contradiction and R/I must be an integral domain and so $I=ker(\varphi)$ is a prime.

Because the two rings are isomorphic, if one has zero divisors then so does the other. Similarly, if one is a field then so is the other. Apply these two facts...

Quote:

Originally Posted by demode

I'm not quiete sure what you mean. But here's what I've tried:

$R/I \cong \mathbb{Z}$, so $(x+I) \mapsto x$. Let $x+I \in R/I$ where

$x+I = \begin{pmatrix}a &3b \\3b & a \end{pmatrix} + I$

Since $a-3b=0 \iff a= 3b$,

$I = \left\{ \begin{pmatrix}3b &3b \\3b & 3b \end{pmatrix} | b \in \mathbb{Z} \right\}$. Therefore

$\varphi \left( \begin{pmatrix}a &3b \\3b & a \end{pmatrix} + I \right) = \varphi \left( \begin{pmatrix}a &3b \\3b & a \end{pmatrix} - \begin{pmatrix}3b &3b \\3b & 3b \end{pmatrix} \right)= \varphi \left( \begin{pmatrix}a-3b &0 \\0 & a-3b \end{pmatrix} + I \right)$

= (a-3b)-(a-3b)=0

Is this also correct?

You want to show that 1 is contained in $im(\varphi)$. That is, does there exist some 2-by-2 matrix of that form which has determinant equal to 1?
• Oct 30th 2010, 09:50 PM
demode
Since the factor ring $R/ ker(\varphi)$ is an integral domain, it follows kernel is a prime ideal. But how do you show that $ker(\varphi)$ is not maximal? More precisely, how do you show that $R/ ker(\varphi)$ is not a field?
• Nov 1st 2010, 01:53 AM
Swlabr
Quote:

Originally Posted by demode
Since the factor ring $R/ ker(\varphi)$ is an integral domain, it follows kernel is a prime ideal. But how do you show that $ker(\varphi)$ is not maximal? More precisely, how do you show that $R/ ker(\varphi)$ is not a field?

You are given that $R/ ker(\varphi) \cong \mathbb{Z}$...