[True or False] Subring of R isomorphic to R/I?

[robertjordan]

Homework Statement

If $I$ is a proper Ideal in a commutative ring $R$, then $R$ has a subring isomorphic to $\frac{R}{I}$.

Book says false...

Homework Equations

$I$ being a proper ideal in $R$ means $I$ is a proper subset of $R$ where
(i) if $a,b\in{R}$ then $a+b\in{R}$
(ii) $0\in{R}$ (Where 0 is the element in R such that for all $r\in{R}$, $r+0=r$
(iii) $a\in{I}$ and $r\in{R}$ implies $ar\in{I}$

$\frac{R}{I}={[a]: a\in{R}}$ where $[a]= {a+r : r\in{R}}$
If $b\in{[a]}$ that means $b\equiv{a}\pmod{I}$ which means $(b-a)\in{I}$
The identity of $\frac{R}{I}$ is $[1]$ where $1$ is the identity of R.

A subring $S$ of $R$ is a subset of $R$ where:
(i) $a,b\in{S}$ implies $a+b\in{S}$ and $ab\in{S}$
(ii) $0\in{S}$ s.t. $a+0=a$ (this is the same 0 as R)
(iii) $1\in{S}$ s.t. $a*1=a$ (this is the same 1 as R)
(iv) for every $a\in{S}$, there is a $(-a)\in{S}$ such that $a+(-a)=0$
(v) $(ab)c=a(bc)$
(vi) $(a+b)+c = a+(b+c)$
(vii) $c*(a+b)= ca+cb$A ring homomorphism from $R$ to $\frac{R}{I}$ is a function $f: R\rightarrow \frac{R}{I}$ such that for all $x,y\in{R}$, $f(xy)=f(x)f(y)$ and $f(x+y)=f(x)+f(y)$ and $f(1_{R})=1_{\frac{R}{I}}$.

The Attempt at a Solution

If $S$ is a subring of $R$, we know by property (iii) that $1\in{S}$.
Now $I$ being a proper ideal of R implies $1\not{\in}{I}$ because if $1\in{I}$ then, for all $r\in{R}$, $r*1\in{I}$ which would mean $I=R$ which is not true because $I$ is a proper ideal.

So if there was an isomorphism $f$ from $S$ to ${R}{I}$, we would have to map the $1$ of $R$ to the $1$ of $\frac{R}{I}$. Well the $1$ of $\frac{R}{I}$ is $[1]$ which is all the elements in $R$ that are congruent to $1$ modulo I. (AKA all the $r\in{R}$ such that $r-1\in{I}$. I was hoping to show this means $1\in{[1]}$ which is a contradiction because $1\not{\in}I$. But perhaps showing that $\frac{R}{I}$ has more elements than $R$ would be a better way to show there can be no isomorphism between them?Thanks

 

[Dick]

Homework Statement

If $I$ is a proper Ideal in a commutative ring $R$, then $R$ has a subring isomorphic to $\frac{R}{I}$.

Book says false...

Homework Equations

$I$ being a proper ideal in $R$ means $I$ is a proper subset of $R$ where
(i) if $a,b\in{R}$ then $a+b\in{R}$
(ii) $0\in{R}$ (Where 0 is the element in R such that for all $r\in{R}$, $r+0=r$
(iii) $a\in{I}$ and $r\in{R}$ implies $ar\in{I}$

$\frac{R}{I}={[a]: a\in{R}}$ where $[a]= {a+r : r\in{R}}$
If $b\in{[a]}$ that means $b\equiv{a}\pmod{I}$ which means $(b-a)\in{I}$
The identity of $\frac{R}{I}$ is $[1]$ where $1$ is the identity of R.

A subring $S$ of $R$ is a subset of $R$ where:
(i) $a,b\in{S}$ implies $a+b\in{S}$ and $ab\in{S}$
(ii) $0\in{S}$ s.t. $a+0=a$ (this is the same 0 as R)
(iii) $1\in{S}$ s.t. $a*1=a$ (this is the same 1 as R)
(iv) for every $a\in{S}$, there is an $a^{-1}\in{S}$ such that $a*a^{-1}=1$
(v) for every $a\in{S}$, there is a $(-a)\in{S}$ such that $a+(-a)=0$
(vi) $(ab)c=a(bc)$
(vii) $(a+b)+c = a+(b+c)$
(viii) $c*(a+b)= ca+cb$

A ring homomorphism from $R$ to $\frac{R}{I}$ is a function $f: R\rightarrow \frac{R}{I}$ such that for all $x,y\in{R}$, $f(xy)=f(x)f(y)$ and $f(x+y)=f(x)+f(y)$ and $f(1_{R})=1_{\frac{R}{I}}$.

The Attempt at a Solution

If $S$ is a subring of $R$, we know by property (iii) that $1\in{S}$.
Now $I$ being a proper ideal of R implies $1\not{\in}{I}$ because if $1\in{I}$ then, for all $r\in{R}$, $r*1\in{I}$ which would mean $I=R$ which is not true because $I$ is a proper ideal.

So if there was an isomorphism $f$ from $S$ to ${R}{I}$, we would have to map the $1$ of $R$ to the $1$ of $\frac{R}{I}$. Well the $1$ of $\frac{R}{I}$ is $[1]$ which is all the elements in $R$ that are congruent to $1$ modulo I. (AKA all the $r\in{R}$ such that $r-1\in{I}$.


I was hoping to show this means $1\in{[1]}$ which is a contradiction because $1\not{\in}I$. But perhaps showing that $\frac{R}{I}$ has more elements than $R$ would be a better way to show there can be no isomorphism between them?


Thanks

I think you've got your definitions somewhat wrong. I've NEVER seen a definition of 'ring' that includes having multiplicative inverses. That looks more like the definition of a field. And ideals that contain 1 are, as you've noted, pretty uninteresting. Could you check those definitions?

 

[robertjordan]

I think you've got your definitions somewhat wrong. I've NEVER seen a definition of 'ring' that includes having multiplicative inverses. That looks more like the definition of a field. And ideals that contain 1 are, as you've noted, pretty uninteresting. Could you check those definitions?

Thanks, I took out the part about commutative rings having multiplicative inverses. I believe the rest of the definitions are correct?

And do you think the right way to go about this problem is to use an argument about the number of elements in $\frac{R}{I}$ when $I$ is a proper ideal versus the number of elements in $R$?

One problem with this is I couldn't find anything in my book about the number of elements in $\frac{R}{I}$


Thanks again

 

[Dick]

Thanks, I took out the part about commutative rings having multiplicative inverses. I believe the rest of the definitions are correct?

And do you think the right way to go about this problem is to use an argument about the number of elements in $\frac{R}{I}$ when $I$ is a proper ideal versus the number of elements in $R$?

One problem with this is I couldn't find anything in my book about the number of elements in $\frac{R}{I}$


Thanks again

That seems pretty ok. There's a few other things but let's not worry about them right now. But if you insist that a subring of a ring with identity 1 must also contain the identity, then you don't want to define an ideal as a subring.

Since you know the book says it's false you should be thinking of a counterexample. Here's the usual example. Take R to be Z, the integers. Then an ideal of Z is the even integers, 2Z, yes? How many elements in the quotient Z/(2Z)?

 

[robertjordan]

That seems pretty ok. There's a few other things but let's not worry about them right now. But if you insist that a subring of a ring with identity 1 must also contain the identity, then you don't want to define an ideal as a subring.

By this do you mean if I insist that subrings of a ring R must have the same identity element as R?

And I agree that proper ideals are not subrings because they don't contain the identity.

Since you know the book says it's false you should be thinking of a counterexample. Here's the usual example. Take R to be Z, the integers. Then an ideal of Z is the even integers, 2Z, yes? How many elements in the quotient Z/(2Z)?

2 elements? Being the equivalence classes of even and odd integers? [0] and [1]...

If that's correct then it will obviously be a problem to find an isomorphism from a ring containing 2 elements to one containing infinite elements.

Is this right?

Thanks so much

 

[Dick]

By this do you mean if I insist that subrings of a ring R must have the same identity element as R?

And I agree that proper ideals are not subrings because they don't contain the identity.
2 elements? Being the equivalence classes of even and odd integers? [0] and [1]...

If that's correct then it will obviously be a problem to find an isomorphism from a ring containing 2 elements to one containing infinite elements.

Is this right?

Thanks so much

Yes, now you are almost there. But an infinite ring can contain a subring consisting of two elements, Z doesn't. Say why.

 

[robertjordan]

Yes, now you are almost there. But an infinite ring can contain a subring consisting of two elements, Z doesn't. Say why.

Any subring S of the integers must contain the identity (the integer 1 in this case), but by existence of additive inverses (which S being a subring gives us) if 1 is in there then -1 is in there, and that means S=$\mathbb{Z}$.

Is this right? Also, out of curiousity, do some books not require subrings to contain the same identity element as the ring they came from?

 

[Dick]

Any subring S of the integers must contain the identity (the integer 1 in this case), but by existence of additive inverses (which S being a subring gives us) if 1 is in there then -1 is in there, and that means S=$\mathbb{Z}$.

Is this right? Also, out of curiousity, do some books not require subrings to contain the same identity element as the ring they came from?

That's one thing. Most books I know define 'ring' without requiring it contain a multiplicative identity at all. A ring that has one is called a 'ring with identity' or 'unit' or something. Z doesn't even include a subring with 2 elements if you don't require an identity. Can you show that?

 

[robertjordan]

That's one thing. Most books I know define 'ring' without requiring it contain a multiplicative identity at all. A ring that has one is called a 'ring with identity' or 'unit' or something. Z doesn't even include a subring with 2 elements if you don't require an identity. Can you show that?

If S is a subring of Z then it either contains a nonzero element in which case it contains all multiples of that element and is of infinite size, or it is just {0} in which case it is size 1.

Is that right?

I was also wondering if you could take a look at my question about PID's? :shy: [edit: nevermind it looks like zodrina has posted a response]

Thanks

 

[Dick]

If S is a subring of Z then it either contains a nonzero element in which case it contains all multiples of that element and is of infinite size, or it is just {0} in which case it is size 1.

Is that right?

I was also wondering if you could take a look at my question about PID's? :shy: [edit: nevermind it looks like zodrina has posted a response]

Thanks

Right exactly. I'll have a look tomorrow. Zondrina is not 100% reliable, so take that advice with a grain of salt.

 

[robertjordan]

Right exactly. I'll have a look tomorrow. Zondrina is not 100% reliable, so take that advice with a grain of salt.

Thank you; I really appreciate your help.

 

[Dick]

Thank you; I really appreciate your help.

No problem. Very welcome. And I see micromass is on the PID thread. micromass is reliable.

 

[wisvuze]

It is false, try R = Z and I = ( 3 ), then R/I is not a subring of Z.

To prove this, remember that a subring must also be a subgroup of the additive abelian group that makes up the ring R.