Modern algebra II, spring 2015.
Practice Quiz 1 solutions
In this quiz we consider only commutative rings.
Mark the boxes that are followed by correct statements.
There are exactly two isomorphism classes of monoids with
2 elements.
True. Such monoid will have 1 and one more element e. For
multiplication there are only two choices ee = e or ee = 1. The
first choice gives us the Elevator monoid, the second - the group
C2 , which is also a monoid.
If e is an idempotent in a ring R, then 1 − e is also an
idempotent.
True by a short computation.
The product e1 e2 of idempotents e1 , e2 in R is an idempotent
in R. (Does the answer change if we don’t assume that R is
commutative?)
True e1 e2 e1 e2 = e1 e1 e2 e2 = e1 e2 . This fail in noncommutative
rings, in general (look for an example in the matrix ring M2 (R)).
Any integral domain is a field.
False, for instance Z is a counterexample.
The set 2 + 4Z is an ideal of Z.
False, any ideal of Z has the form (n) for some n ∈ N.
If a, b are in ideal I of R, their difference a − b is also in I.
True
Any ideal of Z is principal.
True
Any ideal of Z/n is principal for any n > 1.
True. Show that, if a homomorphism R −→ S is surjective and
any ideal of R is principal then any ideal of S is principal as
well.
Any ideal of Z/n × Z/m is principal for any n, m > 1.
True. First show that any ideal of R × S has the form I × J
where I is an ideal of R and J an ideal of S.
Rings Z/24 and Z/6 × Z/4 are isomorphic.
False. Even the underlying abelian groups C24 and C6 × C4 are
not isomorphic.
The set of polynomials without constant term is an ideal of
Q[x].
True. This ideal is (x).
The set of polynomials whose coefficients sum up to zero is
an ideal of Q[x].
True. This ideal is (x − 1).
The set of polynomials with constant term an integer divisible by 3 is an ideal of Z[x].
True. This ideal is (3, x).
The set of polynomials of degree at most 3 is an ideal of
Q[x].
False.
The intersection I ∩ J of ideals I, J of ring R is an ideal of
R.
True
The union I ∪ J of ideals I, J of R is an ideal of R.
False. To get an ideal, form the sum I + J.
A field F has only two ideals.
True
There exists a ring with exactly three ideals.
True, for instance, Z/4.
Automorphisms of a ring constitute a group.
True. Automorphisms of any structure constitute a group.
(this one is hard) Any monoid with 3 elements is commutative.
False. Write a 4-element monoid as {1, a, b}. The only chance
at noncommutativity is if ab 6= ba. Then ab, ba 6= 1, otherwise
they are inverses of each other and the monoid is the group C3 .
So you can try, for instance
ab = a, ba = b.
What about aa and bb? You can try various possibilities, a
simple solution is to declare that any product a1 a2 . . . an where
each ai is either a or b is equal to a1 (the leftmost term). This
gives you an associative product and a monoid (check). This
monoid is noncommutative. Problem: can you find this monoid
in real life, similar to our examples of the Elevator monoid and
the Champaigne monoid?