# ABSTRACT ALGEBRA I EXAM #3

**Problem 1. **Let F be a field. For a polynomial
we denote by

the formal derivative of f.

(a) Let

(b) Let

Is J an ideal of F[x]? Prove or disprove.

Solution. A subset is
an ideal if I is a subgroup under + and is closed under multiplication

by R.

For (a), we clearly have 0 ∈ I so that
if f, g∈I then (f + g)(1) = f(1) + g(1) = 0
and

(f+g)'(1) = f'(1)+g'(1) = 0 so f+g∈I, and if f∈I then (−f)(1) = −f(1) = 0 and
(−f)'(1) = −f'(1) = 0

so −f∈I, hence I is a subgroup under +. Next, if f∈I and p∈R, then (pf)(1) =
p(1)f(1) = 0 and

Thus I is an ideal. Indeed, I is the principal ideal generated by (x − 1)^2.

For (b), we note that J is not an ideal: we have

**Problem 2.** Determine explicitly if the matrix

is a zerodivisor.

**Solution.** The matrix A is a zerodivisor. One can do
this by solving linear equations over Z/26Z directly,

but here is another approach. We note that for a, b, c, d ∈R for any ring R, we
have

where We compute that det(A) = 2 − 15 = (mod 26), when

From this argument, it is easy to see that
if and only if
, and otherwise

is a zerodivisor if and only if det(A) is a zerodivisor.

**Problem 3.** Let R be a ring.

(a) Let a∈R and suppose that
for some .
Show that

(b) Suppose that Show that R is a field.

Solution. Part (a) is easy: if
by the uniqueness of left and

right inverse, we have (You may assume this,
but for completeness: If ab = ca = 1 in a ring R, then

b = c. Indeed, since ab = 1, we have b = cab = c.)

For part (b), we note that by part (a) every
is a unit, so we need only show that R is

commutative. Let a, b∈R, then by hypothesis
then multiplying by a, b on the left and

right, respectively, we have so R is a
field. Such fields exist, e.g. R = Z/2Z and R = Z/3Z.

**Problem 4.** Let R be an integral domain, and let a,
b∈R. Prove that (a) = (b) if and only if a = ub for

some

Solution. We recall that
if and only if b∈(a) if and only if b = ra
for some r∈R. Thus (a) = (b)

if and only if b = ra and a = sb for some r, s∈R. Putting these equations
together, we obtain b = (rs)b, or

b(1−rs) = 0. But since R is an integral domain, this implies that either b = 0
or rs = 1; in the former case,

we then have (b) = (0) = (a) so a = 1b = 0 in which case the result is trivially
true, and in the latter case

we have a = sb where now as claimed.

**Problem 5.** Show that the ideal of Z[i] generated by
2 + i is maximal.

Solution. An ideal I of a commutative ring R is maximal if
and only if R/I is a field. We prove that in

fact which is a field since 5 is prime. We
examine the set of cosets S = Z[i]/(2 + i)

and ask ourselves: what are the possible remainders? What does it mean to
consider elements of Z[i]

“modulo the ideal (2 + i)”? Note that if a + bi + (2 + i)∈S, then a + bi = a +
bi − b(2 + i) = a − 2b so

a+bi+(2+i) = a−2b+(2+i); but also, (2+i)(2−i) = N(2+i) = 5, so we may reduce
a−2b modulo 5

as well. Therefore we define a map

Check that this map is a ring homomorphism. It is
obviously surjective, and its kernel by the above is the

ideal (2 + i). The result then follows.

Alternatively, one can “divide” an element a + bi by 2 + i,
i.e. solve the equation (x + yi)(2 + i) = a + bi;

one obtains

and so x, y∈Z if and only if
(mod 5) (equivalently,
(mod 5)). (This also reproves that

the kernel of the above map is the ideal 2 + i.) Now consider an ideal
then

N(a + bi) = c∈J. If gcd(c, 5) = 1, then 1∈J so J = Z[i], a contradiction.
Therefore must

be a multiple of 5, which can happen if and only if a 2b (mod 5) by a direct
calculation, which is again a

contradiction.