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A ring is a field iff the only ideals are (0)(0) and (1)(1)

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Let RR a ring (with 1∈R1\in R). I want to show that RR is a field ⟺\iff it has exactly two ideals.

For ⟹\implies it’s obvious. Indeed, if RR is a field and I≠{0}I\neq \{0\} an ideal, then, I=AI=A since x∈I⟹xx−1∈Ix\in I\implies xx^{-1}\in I and thus 1∈I1\in I. In particular, if x∈Ax\in A, then x⋅1∈Ix\cdot 1\in I and thus x∈Ix\in I.

For the converse I don’t really know how to do. My idea was the following one:

I show that if RR is not a field then it has more that two ideals. Indeed, let RR be not a field. Then, there is x∈Rx\in R that has no inverse. In particular, (x)∉{R,{0}}(x)\notin \{R,\{0\}\}. Indeed, if (x)=(0)(x)=(0) then x=0x=0 which is a contradiction. If (x)=R(x)=R, then 1∈(x)1\in (x) and thus there is nn s.t. xn=1x^n=1 what prove that xx is invertible and is also a contradiction.

Is it correct ? Is there a more direct proof ? i.e. like this:

Let AA and {0}\{0\} the two ideal of RR. How can I show that all element of A∖{0}A\backslash \{0\} are invertible ? By contradiction, if not, the xy≠1xy\neq 1 for all y∈Ry\in R. Therefore, (x)(x) doesn’t contain 11, and thus, RR has at least three ideal, which is a contradiction.

Is it correct ?

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2 Answers

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Your proof is essentially correct. I would have only stated it a bit differently (and more succintly), as follows.

Let 0≠x∈R0 \ne x \in R and consider the ideal (x)(x). Clearly, (x)≠0(x) \ne 0, therefore necessarily (x)=R(x) = R. This means that there exist y∈Ry \in R such that yx=1yx = 1. There also exist z∈Rz \in R such that xz=1xz = 1 (because (x)(x) is a bilateral ideal). Multiplying yx=1yx = 1 by zz on the right, we get y(xz)=zy(xz) = z, i.e. y=zy = z, so the left and right inverses coincide, therefore xx is invertible. We have therefore proven that every non-zero element is inversible, so RR is a field.

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This assumes commutativity. For example in the matrix ring M2(K)M_2(K), KK a field, the ideal generated by an element x=(0100)x=\pmatrix{0&1\cr0&0\cr} is the entire ring. Yet xx has neither left not right inverse. In fact, M2(K)M_2(K) has only the two trivial ideals, yet it is not a field. The claim is wrong without the commutativity assumption.

– Jyrki Lahtonen♦

2 days ago

@JyrkiLahtonen: You are right, but this remark had already been made by G. Sasatelli before I wrote my answer (see the comments below the question), so I haven’t included it in my post.

– Alex M.

2 days ago

Oops. I did not check the time-stamps, sorry.

– Jyrki Lahtonen♦

2 days ago

Your proof is incorrect. By a small detail, but important:

If (x)=R(x)=R, then 1∈(x)1\in(x) and so there exists y∈Ry\in R with xy=1xy=1.

Maybe your writing of xn=1x^n=1 is just a typo, butâ€¦

You are also assuming RR is commutative, otherwise the statement is false. There are (noncommutative) rings having exactly two (two-sided) ideals, but are not even division rings. For instance the matrix rings over a field (or a division ring, more generally).