Thread: Field and ideals question

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

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).

Why is it that since I is an ideal 1 = (1/x)(x) is also in I?

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).

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.