Ideals of fields

[Dogtanian]

I need to show that the only ideals of a field are the trivial one {0} and f itself.

 

I'm not sure where to begin, or if there are any little tricks to use.

 

So far I've thought about picking an ideal and assume it has an element not equal to 0. (If it doesn't have such an element then it is {0} an we are OK.)

 

With this non-zero element of the ideal I would then add it to a non-zero element of the field. This too is in the ideal. (Am I right in assuming that any ideal of ring an hence a field is a subring under addition?)

 

Since a field has inverses I could pick such an element out of it above so r+f=1, where r is in the ideal and f in the field.

 

Then we have 1 in the ideal. 1 generates the whole field. Thus the ideal is the whole field.

 

 

Is this correct or can you spot any mistakes in it?

 

I also have to show a commutative ring, R, whose only ideals are {0} and R is in fact a field....but I have had a go at that question yet.

[matt grime]

You may not add elements to an element in the ideal like that and expect it to work.

 

An ideal is closed only under adding elements *within* the ideal.

 

It is however closed under multiplication by *ANY* element.

 

As you correctly point out, if the ideal is not 0, then it contains some element x, and more importantly every element in a field has a *MULTIPLICATIVE* inverse.

 

Just by adding -r (which you've called f) above you're not even staying in the ideal, nor are you using the fact that the underlying ring is also a field. So your proof is wrong (and in fact would show that no ring had ideals).

 

 

As for the second question. If R is not a field, then there is some element without a multiplicative inverse. Look at the ideal it generates.

[Dogtanian]

Thanks for your hints...I always get confused over closure of addition and multiplication in ideals...

 

How do these sound now?

 

(1)

Let F be a field, I an ideal of F such that there exists a non-zero r in I. (If there doesn't, then I={0} and we are done).

 

For this r we have a multiplicative inverse, s in F such that sr=1=rs.

 

Since I is closed under multiplication by elements of F 1=sr is in I.

 

Since addition in an ideal is closed 1+1+...+1 is in I.

 

Any element in F can be written as 1+1+...+1 for some number of 1s, thus any element of F is in I.

By definition, any element of I is in F, thus the two are equal.

 

 

(2)

Let R be a commutative ring whose only ideals are {0} and R.

 

Suppose R is not a field.

Then there exists r in R such that there is no multiplicative inverse for r in R.

 

There is an ideal generated by r; <r> = { pr : pis in R} (={rp : p is in R} since R commutative).

 

So we have <r> = {0} or R (by definition of R).

 

<r> = {0} iff r = 0, so suppose ris not 0.

Then <r> must be R.

 

But we have 1 in R, which implies that pr=1=rp for some p in R.

That is, there exists a multiplicative inverse for r in R.

 

This is a contadiction and hence R must be a field.

[matt grime]

Thanks for your hints...I always get confused over closure of addition and multiplication in ideals...

 

How do these sound now?

 

(1)

Let F be a field' date=' I an ideal of F such that there exists a non-zero r in I. (If there doesn't, then I={0} and we are done).

 

For this r we have a multiplicative inverse, s in F such that sr=1=rs.

 

Since I is closed under multiplication by elements of F 1=sr is in I.[/quote']

 

 

no, the ideal contains 1 and since it is an ideal it contains 1.x for all x.... what you're about to continue with here -

 

Since addition in an ideal is closed 1+1+...+1 is in I.

 

Any element in F can be written as 1+1+...+1 for some number of 1s, thus any element of F is in I.

By definition, any element of I is in F, thus the two are equal.

 

not useful, nor indeed correct since not every element in a field or ring is obtained by adding 1 to itself (only the subring of integers at most is obtained).

 

(2)

Let R be a commutative ring whose only ideals are {0} and R.

 

Suppose R is not a field.

Then there exists r in R such that there is no multiplicative inverse for r in R.

 

There is an ideal generated by r; <r> = { pr : pis in R} (={rp : p is in R} since R commutative).

 

So we have <r> = {0} or R (by definition of R).

 

<r> = {0} iff r = 0, so suppose ris not 0.

Then <r> must be R.

 

But we have 1 in R, which implies that pr=1=rp for some p in R.

That is, there exists a multiplicative inverse for r in R.

 

This is a contadiction and hence R must be a field.

 

yes.

[Dogtanian]

Thanks. I think I get it now