Try the Free Math Solver or Scroll down to Resources!












Please use this form if you would like
to have this math solver on your website,
free of charge.


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

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