Commutative Ring with Unique Ideals: Solution Using \mathbb{Z}_4

[noospace]

Homework Statement

I'm trying to find a commutative ring, not a field, who's only ideals are {0} and itself.

Homework Equations

Definition: A subset of a ring R is an ideal if it is a subring of R and is closed under multiplication by elements of R.

The Attempt at a Solution

I claim $\mathbb{Z}_4$ has the desired properties.
$\mathbb{Z}_4$ is a commutative ring.
$\mathbb{Z}_4$ is not a field since it has zero divisors.
The only proper nontrivial subring of $\mathbb{Z}_4$ is {0,2} and {0,2} is not an ideal (not closed under multiplication by 2).

 

[d_leet]

Homework Statement

I'm trying to find a commutative ring, not a field, who's only ideals are {0} and itself.

Was this actually the question? Or was it something more of the form "does there exist a commutative ring whose only ideals are {0} and itself that is not a field?"

Definition: A subset of a ring R is an ideal if it is a subring of R and is closed under multiplication by elements of R.

This definition isn't quite right. An ideal, I, is a subring such that rI and Ir are subsets of I for every r in R. (for a two-sided ideal).

I claim $\mathbb{Z}_4$ has the desired properties.
$\mathbb{Z}_4$ is a commutative ring.
$\mathbb{Z}_4$ is not a field since it has zero divisors.
The only proper nontrivial subring of $\mathbb{Z}_4$ is {0,2} and {0,2} is not an ideal (not closed under multiplication by 2).

But if you multiply the elements in {0,2} by 2 you get {0} which is contained in {0,2}.

If the condition were that rI=I for every r in the ring than there would only be one ideal {0}, because 0 must be in the ideal since it is an additive group and multiplying by 0 will only give zero then 0*I would be {0}.

 

[noospace]

Was this actually the question? Or was it something more of the form "does there exist a commutative ring whose only ideals are {0} and itself that is not a field?"



This definition isn't quite right. An ideal, I, is a subring such that rI and Ir are subsets of I for every r in R. (for a two-sided ideal).



But if you multiply the elements in {0,2} by 2 you get {0} which is contained in {0,2}.

If the condition were that rI=I for every r in the ring than there would only be one ideal {0}, because 0 must be in the ideal since it is an additive group and multiplying by 0 will only give zero then 0*I would be {0}.

You're correct and in fact $Z_4$ does not satisfy the required properties since {0,2} is not an ideal.

 

[matt grime]

{0,2} is an ideal of Z_4: there is at least one too many negatives in your last sentence.

Look, let's just write down what we know:

suppose we have a ring R and the only ideals in R are 0 and R.

Let's suppose that x is a non-identity element in R and consider the ideal xR, since 1R=R, and this tells us nothing about R.

xR is either 0 or R.

Since rings normally assumed to have identities you can now proceed to deduce some things.

 

[noospace]

{0,2} is an ideal of Z_4: there is at least one too many negatives in your last sentence.

Look, let's just write down what we know:

suppose we have a ring R and the only ideals in R are 0 and R.

Let's suppose that x is a non-identity element in R and consider the ideal xR, since 1R=R, and this tells us nothing about R.

xR is either 0 or R.

Since rings normally assumed to have identities you can now proceed to deduce some things.

My bad again. I meant to say that Z4 IS an ideal. Thanks.

Am I missing something or does the trivial ring R = {0} not count??

R is a commutative ring.
R is not a field since it does not have a nonzero identity.
The only subrings of R are {0} and R.
Weird.

 

[matt grime]

Most people require a ring to have a multiplicative identity. Why is it weird that there are (degenerate) counter examples. If you follow my suggestion you will find all of the counter examples.