Misc. Algebra

[tirix]

Additional theorems which should be useful for #4246.
14 theorems moved to main, mainly (sub)division rings related theorems from @icecream17

[icecream17]

a bit faster review than normal

[langgerard]

Another interesting characterisation of division rings is that they only have two ideals ( the 0 ideal and the ring itself)
because if a is non zero in the division ring R, the multiplication with its inverse is in any non-zero ideal I, so that 1 is in the ideal I and I=R.
Now if the ring R has only two ideals, then if I is the non zero ideal, it must be that I =A, so that 1 is in I.
If a is non-zero, it must have a left inverse r such that r.a = 1. But also r is non-zero, and must have a left inverse s,
so that 1 = s.r and we have s = s.(r.a)=(s.r).a = a so that also s.r = a.r = 1, proving that r is the bilateral inverse of a and that every non-zero element of A is inversible, so that R is a division ring.
By the way, the zero ring is the only ring with only one ideal !

[langgerard]

Another interesting characterisation of division ring is that they are the only rings with two ideals (every ring R must atv least have two ideals 0 and R, except a zero ring where 0 = R). If R is a division ring and a is nonzero in the non zero ideal I, then the inverse element b of a is such thatv a.b (or b.a) is in I, so that1 is in I, and so I = A.
If I is the only non zero ideal in the non zero ring R, then I = R and if a is a nonzero element in I, because 1 is in I, there exist an element r in R such that r.a =1, so that r is a left inverse of a. But also there is an element s in R such that s.r= 1., and s is a left inverse of r. Then we have s = s.(r.a) = (s.r).a = a, proving that s = a and r is a bilateral inverse of a.
As every non ero element in R is inversible, R must ne a division ring

[tirix]

Another interesting characterisation of division rings is that they only have two ideals ( the 0 ideal and the ring itself)
because if a is non zero in the division ring R, the multiplication with its inverse is in any non-zero ideal I, so that 1 is in the ideal I and I=R.
Now if the ring R has only two ideals, then if I is the non zero ideal, it must be that I =A, so that 1 is in I.
If a is non-zero, it must have a left inverse r such that r.a = 1. But also r is non-zero, and must have a left inverse s,
so that 1 = s.r and we have s = s.(r.a)=(s.r).a = a so that also s.r = a.r = 1, proving that r is the bilateral inverse of a and that every non-zero element of A is inversible, so that R is a division ring.
By the way, the zero ring is the only ring with only one ideal !

This would be ~ drngidl !

In my next PR, I plan to prove that the quotient of a (commutative) ring by a maximal ideal is a division ring.
The two proofs seem very similar : is there a way to use one to prove the other?

[langgerard]

In the case that R is a ring and M is a maximal bilateral ideal in R, then the quotient ring R/M has only two ideals 0 and R/M itself, so that R/M must be a division ring. I think that it works even if R is not commutative.

[langgerard]

And now suppose that N is a left maximal ideal in R, then M C N, and if N strictly includes N, let x be an element in N/M.
Then the class x* of x in R/M is not 0, so that R/M being a division ring, there is aonther class y* in R/M with x*.y* = 1 = y*.x*.
So that there will be an element y in M/N (because y* cannot be 0) such that x.y = 1 =y.x. But then 1 will be in the ideal N, and we have N = R and N will be bilateral.
This proves that if R is a ring and M a left( (or right) maximal ideal, then R/M will also be a division ring

[langgerard]

Easier : Let N be a left ideal between M and R, then the projection ofN into R/M must be an ideal in R/M, so must be R or M. So that there can be no left (or right) maximal ideal of R strictly including M. And this proves that if R is a (unital) ring and M an unilateral or bilateral maximal ideal in R, then R/M must be a division ring

[benjub]

It may be easier in the other direction:

Let $R$ be a ring and $M$ be a proper ideal of $R$ which is maximal both as a left-ideal and as a right-ideal.
Let $a \in R/M$ be nonzero, so $a = [x]$ with $x \in R \setminus M$.
Since $M$ is maximal as a left-ideal, the left-ideal generated by $M$ and $x$ is the whole $R$, so there exists $r \in R$ and $m \in M$ such that $rx + m = 1$, so $[r][x] = 1$. So $a$ is left-invertible.
Same on the right.
Therefore, $R/M$ is a division ring.

Now, suppose that a nonzero ring $R$ has at most two left-ideals and at most two right-ideals (which are necessarily {0} and $M$ since these are ideals). Then, {0} is maximal both as a left-ideal and as a right-ideal, so $R/${0} = $R$ is a division ring by the previous result.

[tirix]

Thanks for all the great suggestions!
I'm merging this PR now, I'll work on them for the next one.