0
No. For instance in R, which is commutative, we have the ideals (2) and (4), where (4) is strictly contained in (2), which is not R. Therefore the ideal (4) is not a maximal ideal.
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
Can one formally define a division ring as a field that isn't necessarily commutative?
wedderburn's little theorem says all finite division rings are commutative so they are fields. So if it is a finite division ring, then the answer is NO But for an infinite division ring... I think you can!
How field is denoted in algebra?
A field is a commutative ring in which all non zero elements have inverses or all the elements are units
Every finite division ring is a field?
correct.
One tree ring equals how much time?
A tree gets a new ring every year, so I suppose a tree ring equals one year.
Three clocks ring once at the same time After that e first clock rings after every 3 hours the second after every 4 hours and third after every 5 hours After how many hours will they again ring?
they never meet again at the same time
Problem and solution for maximal ideal in mathematics?
Given a ring R and a proper ideal I of R(that is I ≠ R), I is called a maximal ideal of R if there exists no other proper ideal J of R so that I ⊂ J. Is that the ideal you are talking about? If so, not sure what you want to know?
How do you prove a Ring to be commutative?
To prove a ring is commutative, one must show that for any two elements of the ring their product does not depend on the order in which you multiply them. For example, if p and q are any two elements of your ring then p*q must equal q*p in order for the ring to be commutative. Note that not every ring is commutative, in some rings p*q does not equal q*p for arbitrary q and p (for example, the ring of 2x2 matrices).
Is Every commutative ring with unity is an integral domain?
Nope. Take the commutative ring Z4 (the set {0,1,2,3} with modular arithmetic and multiplication). 1 is the identity element. But 2 x 2 = 4 = 0 while 2 is not the 0 element. So it's not an integral domain.
Can one formally define a division ring as a field that isn't necessarily commutative?
wedderburn's little theorem says all finite division rings are commutative so they are fields. So if it is a finite division ring, then the answer is NO But for an infinite division ring... I think you can!
What does it mean to have a non example of divisor?
A non-example of divisor ring of integers, a division ring or a nonzero commutative ring that has no zero divisors except 0.
What does maximal set of ring mean?
If A intersection E = A where A & E belongs to some some system of sets lets say S then E is called the maximal set of S or the unit set of S.
What has the author David Dobbs written?
David Dobbs has written: 'Advances in Commutative Ring Theory'
What do you know about whole number?
They form a commutative ring in which the primary operator is addition and the secondary operator is multiplication.
How field is denoted in algebra?
A field is a commutative ring in which all non zero elements have inverses or all the elements are units
What does an OH group at position 3 and 4 of the benzene ring of adrenergic drugs result in?
It results in maximal alpha and beta receptor activity.
Can you use your regular ring as a promise ring The ring is a christian ring with scriptures on it Want to no if it can be used as a promise ring Can really afford a real one?
I think that such a 'Promise' ring would be an ideal gift .
How you should know about whole number?
They form a commutative ring in which the primary operation is addition and the secondary operation is multiplication. However, it is not a field because it is not closed under division by a non-zero element.
