Ideals of a Residue Class Ring- Ring Isomorphism

[Math Amateur]

I am reading R. Y. Sharp: Steps in Commutative Algebra.

In Chapter 2: Ideals on page 32 we find Exercise 2.40 which reads as follows:

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Let I, J be ideals of the commutative ring R such that \(\displaystyle I \subseteq J \).

Show that there is a ring isomorphism

\(\displaystyle \xi \ : (R/I) \ / \ (J/I) \to R/J \)

for which

\(\displaystyle \xi ((r + I) + J/I ) = r + J \) for all \(\displaystyle r \in R\).

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Can someone please help me get started on this exercise.

Also ... problem ... considering J/I ... for a factor ring, J is usually a ring, so what does J/I mean? I assume that since ideals are subgroups under addition, then we can make sense of this by interpreting J/I as a factor group ... Is this the case ... can someone please confirm that my view on this matter is valid ...

As indicated above i would be grateful for some help to get started ...

Peter

 

[Deveno]

This is also known as "The Second Isomorphism Theorem" and is a direct analogue of a similar result for groups.

If $J$ is a subring of $R$ containing an ideal $I$ of $R$, it is clear to see that $I$ is also an ideal of $J$. So, yes, $J/I$ is just a "smaller" factor ring, and it should be clear that it is a subring of $R/I$.

You see, every ideal IS a subring (if one stipulates that subrings need not contain unity):

Recall that a subring has to be closed under multiplication. Well if $a\in J$, and $x \in J$, then a fortiori (loosely translated: "all the more so", literally, "from the stronger" an abbreviation of "a fortiori argumento" -from the stronger argument), $a \in R$, and since $J$ is an ideal, $ax \in J$, hence $J$ is closed under multiplication.

However, a subring need not be an ideal: consider the ring $\Bbb Z \times \Bbb Z$ with the multiplication:

$(a,b)(a',b') = (aa',bb')$.

The set $\Delta \Bbb Z = \{(k,k)\}$ is a subring of $\Bbb Z \times \Bbb Z$, but it is NOT an ideal; Let $a,k \neq 0$, then:

$(a,0)(k,k) = (ak,0) \not \in \Delta \Bbb Z$.

Now, we are already given a definition of $\xi$, but as with any mapping defined on a quotient ring via elements of the ring, we must ensure that it is "well-defined", i.e; that it is constant on any coset. Because $\xi$ is defined on a "coset of a coset" we have to do this TWICE:

First: suppose $(r + I) + J/I = (r' + I) + J/I$ (as cosets in $(R/I)/(J/I)$).

This means that $(r + I) - (r' + I) \in J/I$

But also: in $J/I$, we have:

$(r + I) - (r' + I) = (r - r') + I$

so if this coset is in $J/I$, we must have that $r - r' \in J$, so $r + J = r' + J$ in $R/J$.

But THAT means that $\xi((r+I) + J/I)) = r + J = r' + J = \xi((r'+I) + J/I)$

whenever $(r + I) + J/I = (r' + I) + J/I$, that is: $\xi$ is well-defined.

It is then trivial that $\xi$ is surjective, since for every $r + J \in R/J$, we have the pre-image:

$(r + I) + J/I$ in $(R/I)/(J/I)$ (use the same $r$).

It, of course, remains to be seen that $\xi$ is a ring-homomorphism, which I leave to you.

I will address one final point, the kernel of $\xi$. This is, by definition:

$\{(r + I) + J/I \in (R/I)/(J/I): \xi((r + I) + J/I) = J\}$.

Since $r + J = J \iff r \in J$, this is the set:

$\{(r + I) + J/I: r \in J\}$, and this means $r+I \in J/I$, whence:

$\text{ker}(\xi) = \{J/I\}$, the 0-element of $J/I$.

Here is a painfully worked-out example:

Let $R = \Bbb Z$ and let $I = (12), J = (3)$.

Using the canonical isomorphism:

$\Bbb Z/(12) \cong \Bbb Z_{12}$ let's look at what $J/I$ is, explicitly.

To simplify the notation, we will use $[k]_{12}$ for the image of $k$ in $\Bbb Z_{12}$

under the canonical isomorphism.

Then $J/I = {[k]_{12} \in \Bbb Z_9: 3|k} = \{[0]_{12},[3]_{12},[6]_{12},[9]_{12}\}$

Then the possible cosets are:

$[0]_{12} + J/I = J/I = \{[0]_{12},[3]_{12},[6]_{12},[9]_{12}\}$
$[1]_{12} + J/I = \{[1]_{12},[4]_{12},[7]_{12},[10]_{12}\}$
$[2]_{12} + J/I = \{[2]_{12},[5]_{12},[8]_{12},[11]_{12}\}$

(we could, if we so chose, simply this notation even further by representing these cosets with double brackets: $[[k]]$, meaning first we reduce $k$ mod 12, and then reduce the resulting "integer" again mod 3 -this works on the entire cosets because 3|12, which is exactly the necessary condition we must have for (12) to be a ideal contained in (3)).

Here, our isomorphism is:

$\xi([k]_{12} + J/I) = [k]_3$

(which makes sense: $J/I$ is a finite subring of order 4, so the quotient should have order 12/4 = 3).

Sometimes this theorem is called the "freshman's theorem": you just "cancel the $I$'s".

 

[Math Amateur]

This is also known as "The Second Isomorphism Theorem" and is a direct analogue of a similar result for groups.

If $J$ is a subring of $R$ containing an ideal $I$ of $R$, it is clear to see that $I$ is also an ideal of $J$. So, yes, $J/I$ is just a "smaller" factor ring, and it should be clear that it is a subring of $R/I$.

You see, every ideal IS a subring (if one stipulates that subrings need not contain unity):

Recall that a subring has to be closed under multiplication. Well if $a\in J$, and $x \in J$, then a fortiori (loosely translated: "all the more so", literally, "from the stronger" an abbreviation of "a fortiori argumento" -from the stronger argument), $a \in R$, and since $J$ is an ideal, $ax \in J$, hence $J$ is closed under multiplication.

However, a subring need not be an ideal: consider the ring $\Bbb Z \times \Bbb Z$ with the multiplication:

$(a,b)(a',b') = (aa',bb')$.

The set $\Delta \Bbb Z = \{(k,k)\}$ is a subring of $\Bbb Z \times \Bbb Z$, but it is NOT an ideal; Let $a,k \neq 0$, then:

$(a,0)(k,k) = (ak,0) \not \in \Delta \Bbb Z$.

Now, we are already given a definition of $\xi$, but as with any mapping defined on a quotient ring via elements of the ring, we must ensure that it is "well-defined", i.e; that it is constant on any coset. Because $\xi$ is defined on a "coset of a coset" we have to do this TWICE:

First: suppose $(r + I) + J/I = (r' + I) + J/I$ (as cosets in $(R/I)/(J/I)$).

This means that $(r + I) - (r' + I) \in J/I$

But also: in $J/I$, we have:

$(r + I) - (r' + I) = (r - r') + I$

so if this coset is in $J/I$, we must have that $r - r' \in J$, so $r + J = r' + J$ in $R/J$.

But THAT means that $\xi((r+I) + J/I)) = r + J = r' + J = \xi((r'+I) + J/I)$

whenever $(r + I) + J/I = (r' + I) + J/I$, that is: $\xi$ is well-defined.

It is then trivial that $\xi$ is surjective, since for every $r + J \in R/J$, we have the pre-image:

$(r + I) + J/I$ in $(R/I)/(J/I)$ (use the same $r$).

It, of course, remains to be seen that $\xi$ is a ring-homomorphism, which I leave to you.

I will address one final point, the kernel of $\xi$. This is, by definition:

$\{(r + I) + J/I \in (R/I)/(J/I): \xi((r + I) + J/I) = J\}$.

Since $r + J = J \iff r \in J$, this is the set:

$\{(r + I) + J/I: r \in J\}$, and this means $r+I \in J/I$, whence:

$\text{ker}(\xi) = \{J/I\}$, the 0-element of $J/I$.

Here is a painfully worked-out example:

Let $R = \Bbb Z$ and let $I = (12), J = (3)$.

Using the canonical isomorphism:

$\Bbb Z/(12) \cong \Bbb Z_{12}$ let's look at what $J/I$ is, explicitly.

To simplify the notation, we will use $[k]_{12}$ for the image of $k$ in $\Bbb Z_{12}$

under the canonical isomorphism.

Then $J/I = {[k]_{12} \in \Bbb Z_9: 3|k} = \{[0]_{12},[3]_{12},[6]_{12},[9]_{12}\}$

Then the possible cosets are:

$[0]_{12} + J/I = J/I = \{[0]_{12},[3]_{12},[6]_{12},[9]_{12}\}$
$[1]_{12} + J/I = \{[1]_{12},[4]_{12},[7]_{12},[10]_{12}\}$
$[2]_{12} + J/I = \{[2]_{12},[5]_{12},[8]_{12},[11]_{12}\}$

(we could, if we so chose, simply this notation even further by representing these cosets with double brackets: $[[k]]$, meaning first we reduce $k$ mod 12, and then reduce the resulting "integer" again mod 3 -this works on the entire cosets because 3|12, which is exactly the necessary condition we must have for (12) to be a ideal contained in (3)).

Here, our isomorphism is:

$\xi([k]_{12} + J/I) = [k]_3$

(which makes sense: $J/I$ is a finite subring of order 4, so the quotient should have order 12/4 = 3).

Sometimes this theorem is called the "freshman's theorem": you just "cancel the $I$'s".

Thanks for the help Deveno!

Just carefully working through this now.

Peter