Prove that a boolean ring r is a field if and only if r contains only 0 and 1
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You are trying to prove the equivalence of the following statements:
P:P: A commutative ring RR with 11 is a field.
Q:Q: The only ideals of RR are (0)(0) and (1)(1).
Let us look at statement QQ closely. Well it is saying that the only ideals of RR are the zero ideal (which has only one element, zero) and (1)(1). What is (1)(1)? Well by definition of an ideal if you multiply anything in the ideal (1)(1) by anything in RR, you should get something back in (1)(1)again. But then 11 is the multiplicative identity of RR so multiplying everything in RR by 11 just gives everything in RR back again. This means that (1)(1) must be the whole ring RR.
Now suppose we want to prove P⟹QP⟹Q. Let II be an ideal of a ring RR. Here "ring" means "commutative ring with a unit". Now here are some things you should know:
(1) II is non-empty
(2) II must at least contain the element 00 (Why?)
(3) If II has more than one element this means that at least one non-zero element aa of the ring must be in II. (Why?)
Therefore if II contains only 00, I=(0)I=(0). If II does not only contain zero, then by (3) above it contains at least one non-zero element aa of RR. Now recall that we are trying to prove P⟹QP⟹Q. We already know PP. Therefore this means by definition of a field that a−1a−1 exists in RR.
But then by definition of an ideal II, a−1a=1a−1a=1 must be in II. Therefore 1∈I1∈Iso that II must be the whole ring RR. Hence I=(1)I=(1). This establishes P⟹QP⟹Q.
Now for the converse:
To show Q⟹PQ⟹P it suffices to show that non-zero every element a∈Ra∈Rcontains a multiplicative inverse. So let aabe a non-zero element of RR. The trick now is to consider the principal ideal generated by aa (which we denote by (a)(a)).
Now by assumption of QQ, since the only ideals of RR are (0)(0) and (1)(1), this means that (a)(a) being an ideal of RR must be either (0)(0) or (1)(1). Now (a)(a) cannot be (0)(0)for a≠0a≠0. So (a)=(1)(a)=(1). But then this means that 11 is a multiple of aa, viz. there exists a non-zero cc such that
ac=1.ac=1.
However this is precisely saying that aahas a multiplicative inverse. Since aa was an arbitrary non-zero element of RR, we are done. Q.E.D.
P:P: A commutative ring RR with 11 is a field.
Q:Q: The only ideals of RR are (0)(0) and (1)(1).
Let us look at statement QQ closely. Well it is saying that the only ideals of RR are the zero ideal (which has only one element, zero) and (1)(1). What is (1)(1)? Well by definition of an ideal if you multiply anything in the ideal (1)(1) by anything in RR, you should get something back in (1)(1)again. But then 11 is the multiplicative identity of RR so multiplying everything in RR by 11 just gives everything in RR back again. This means that (1)(1) must be the whole ring RR.
Now suppose we want to prove P⟹QP⟹Q. Let II be an ideal of a ring RR. Here "ring" means "commutative ring with a unit". Now here are some things you should know:
(1) II is non-empty
(2) II must at least contain the element 00 (Why?)
(3) If II has more than one element this means that at least one non-zero element aa of the ring must be in II. (Why?)
Therefore if II contains only 00, I=(0)I=(0). If II does not only contain zero, then by (3) above it contains at least one non-zero element aa of RR. Now recall that we are trying to prove P⟹QP⟹Q. We already know PP. Therefore this means by definition of a field that a−1a−1 exists in RR.
But then by definition of an ideal II, a−1a=1a−1a=1 must be in II. Therefore 1∈I1∈Iso that II must be the whole ring RR. Hence I=(1)I=(1). This establishes P⟹QP⟹Q.
Now for the converse:
To show Q⟹PQ⟹P it suffices to show that non-zero every element a∈Ra∈Rcontains a multiplicative inverse. So let aabe a non-zero element of RR. The trick now is to consider the principal ideal generated by aa (which we denote by (a)(a)).
Now by assumption of QQ, since the only ideals of RR are (0)(0) and (1)(1), this means that (a)(a) being an ideal of RR must be either (0)(0) or (1)(1). Now (a)(a) cannot be (0)(0)for a≠0a≠0. So (a)=(1)(a)=(1). But then this means that 11 is a multiple of aa, viz. there exists a non-zero cc such that
ac=1.ac=1.
However this is precisely saying that aahas a multiplicative inverse. Since aa was an arbitrary non-zero element of RR, we are done. Q.E.D.
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