Can someone explain why a field can have no non-trivial ideals?

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- Nov 14th 2012, 09:06 PMjzelltField and ideals question
Can someone explain why a field can have no non-trivial ideals?

- Nov 14th 2012, 11:42 PMDevenoRe: Field and ideals question
suppose we have a field F (which is, of course, also a ring). let I be an ideal of F that contains x ≠ 0.

since x ≠ 0, we have 1/x in F, and since I is an ideal 1 = (1/x)(x) is also in I.

thus for ANY a in F, a = (a)(1) is in I, so that I = F.

(in any ring with identity, if an ideal I contains 1, it is the entire ring). - Nov 14th 2012, 11:59 PMjzelltRe: Field and ideals question
Why is it that since I is an ideal 1 = (1/x)(x) is also in I?

- Nov 15th 2012, 09:03 AMDevenoRe: Field and ideals question
for an ideal,I, of a ring R, we have:

for any x in I, and all r in R: rx is in I (ideals "absorb" elements when you multiply by them).

an ideal is an additive subgroup that "acts like 0":

I + I = I

-I = I

RI = IR = I

just like:

0 + 0 = 0

-0 = 0

a0 = 0a = 0

it is these "zero-like" properties that allow ideals to serve as "ring-kernels" (when we form a quotient ring R/I, we are essentially "setting all of I equivalent to 0" since I is the additive identity of R/I). - Feb 12th 2013, 11:13 PMarulanandamRe: Field and ideals question
Can someone explain why a field can have no non-trivial ideals?

Ans:

If R is a field and U is any nontrivial ideal of R then 1 belongs to U then by theorem U = R for any Ideal U of the field R.