If R does not have an identity, the statement is not true. For instance, consider R = 2Z, the ring of all even integers, and M = 4Z, the ring of all integer multiples of 4. M is a maximal ideal in R, but if we let a=6, a∉M but aR = 12Z ⊂ M, so M+aR = M ⊂ R.
If R is a ring with identity, you only need to prove that M+aR is an ideal; then since M is maximal and a ∈ M+aR \ M (this is where we need the identity: to prove that a ∈ M+aR), it follows that M+aR = R.
If R is commutative we can prove this as follows: Let k ∈ R and let m + ar ∈ M+aR s.t. m ∈ M.
Then k(m+ar) = km + kar = km + a(kr); since M is an ideal, km ∈ M and hence k(m+ar) ∈ M+aR. So M+aR is an ideal.
If R is not commutative, we can still prove that M+aR is a right ideal of R:
Let k, m, a, r be as before. Then (m+ar)k = mk + (ar)k = mk + a(rk); since mk ∈ M and rk ∈ R we have (m+ar)k ∈ M+aR. So M+aR is a right ideal of R.
However, I don't see a way to prove that M+aR is a left ideal of R for a non-commutative ring, and I suspect the statement may well be false for non-commutative rings in general, though I can't find a counterexample either.
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If R does not have an identity, the statement is not true. For instance, consider R = 2Z, the ring of all even integers, and M = 4Z, the ring of all integer multiples of 4. M is a maximal ideal in R, but if we let a=6, a∉M but aR = 12Z ⊂ M, so M+aR = M ⊂ R.
If R is a ring with identity, you only need to prove that M+aR is an ideal; then since M is maximal and a ∈ M+aR \ M (this is where we need the identity: to prove that a ∈ M+aR), it follows that M+aR = R.
If R is commutative we can prove this as follows: Let k ∈ R and let m + ar ∈ M+aR s.t. m ∈ M.
Then k(m+ar) = km + kar = km + a(kr); since M is an ideal, km ∈ M and hence k(m+ar) ∈ M+aR. So M+aR is an ideal.
If R is not commutative, we can still prove that M+aR is a right ideal of R:
Let k, m, a, r be as before. Then (m+ar)k = mk + (ar)k = mk + a(rk); since mk ∈ M and rk ∈ R we have (m+ar)k ∈ M+aR. So M+aR is a right ideal of R.
However, I don't see a way to prove that M+aR is a left ideal of R for a non-commutative ring, and I suspect the statement may well be false for non-commutative rings in general, though I can't find a counterexample either.
If R is a ring with identity (possibly non-commutative), M is a maximal 2-sided ideal, and a
is not in M, then M + RaR is a two sided ideal strictly containing M. Therefore
M + RaR = R.
Comments: Let f : R --> R/M = the quotient homomorphism onto the quotient ring R/M.
R/M has no proper 2 sided ideals except (0). Then since a is not a member of M, f(a) = a'
is non-0. Therefore R/M is a proper 2-sided ideal generated by a'.
If R is commutative then R/M is a field, because, for every non-0 x in R/M , the ideal
x [R/M] = R/M so there exists some y in R/M with xy = I . Therefore x is invertible.
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Similarly, if M is a maximal right ideal in R not containing a, M + aR = R.
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Next consider M_2 = the ring of 2x2 matrices with real entries, which is isomorphic to
the ring of linear transformations on 2 dimensional space.
(0) is the only proper 2-sided ideal in M_2, so it is maximal.
Let A be a linear transformation with 1 dimensional range -- for example
Ae1 = e1, Ae2 = 0,
where e1 & e2 are the standard basis vectors of R^2. The range of A is spanned by e1.
(0) + AM_2 = AM_2 = { AT: T is in M_2 } . The range of AT is contained in the range of A,
so AM_2 =/= M_2 .
Likewise the left ideal M_2 A =/= M_2
This example verifies Scarlet Manuka's conjecture.