Exercises on ring homomorphisms

  1. Exercise 1. Consider the application

    φ: ℤ ⟶ ℤ

      z ⟼ kz

    Determine for which values of k ∈ ℤ, φ is a ring homomorphism.

  2. Exercise 2. Prove that in the ring R of n x n matrix with element in a field K, the subset of matrices with the last row (column) made by zero elements is a right (left) ideal, but not a two-sided ideal.

  3. Exercise 3. Let R the ring of n x n matrix with real elements and let S the subset of R of the singular matrix (that is with null determinant). Determine if S is an ideal and if left, right or two-sided of R.

  4. Exercise 4. Let A the ring of real continuos functions defined in the interval [-1,1]. Let S = {f ∈ A | f(0) = 7} and T = {f ∈ A | f(1/7) = 0}. Verify if S and T are ideals or rings or neither.

  5. Exercise 5. Prove that a field 𝕂 posseses only the trivial ideals {0} and 𝕂 itself.

  6. Exercise 6. Let X a subset of a ring R. Define the following subsets

    A(X) := {r ∈ R | rx = 0 ∀x ∈ X}

    B(X) := {r ∈ R | xr = 0 ∀x ∈ X}

    prove that A(X) is a left ideal of R, B(x) is a right ideal of R. If X is a left ideal of R prove that A(X) is a two-sided ideal.

  7. Exercise 7. Let R = M2(ℝ) the ring of 2 × 2 matrix with real elements. Let X = {(x,0; y 0) | x,y ∈ ℝ}. With the same sets defined in the previous exercise determine A(X) and B(x). Indicate if they are two-sided ideals.

  8. Exercise 8. An element a of a ring R is called nilpotent if am = 0 for some positive integer m. Prove that the set N of all nilpotent elements of a commutative ring R is an ideal of R. Show the commutative requirement is necessary.

  9. Exercise 9.

    Let a0 an element of a field 𝕂 and let R a ring 𝕂[x]. Determine the values a0 for which the subset R of the polynomials which have constant term a0 is an ideal of R.

  10. Exercise 10. In the ring ℚ[x] consider the following subset

    I = {f(x) (x2 + 5) | f(x) ∈ ℚ[x]}

    J = {f(x) (x2 −x + 1) | f(x) ∈ ℚ[x]}

    Prove that I and J are ideals in ℚ[x] and express the quotient rings of ℚ[x]/I and ℚ[x]/J.

  11. Exercise 11. Let R a commutative ring and N the set all nilpotent elements of R. We already showed that it is an ideal of R. Prove that the quotient R/N does not possess non-null nilpotent elements.

  12. Exercise 12. Let I and J two two-sided ideals of a ring R. Let

    I + J := {a + b | a ∈ I, b ∈ J}   IJ := {∑ni=0 aibi | n ∈ ℕ, ai ∈ I, bi ∈ J}

    Prove that I + J and IJ are two-sided ideals of R.

  13. Exercise 13. Let R a commutative ring with unit. Let I and J two ideals of R such that I + J = R. Prove that IJ = I ∩ J.

  14. Exercise 14. Find two nilpotent 2×2-matrices A and B with A+B not nilpotent, you can even have A+B invertible.

    π−1(π(A)) = A + I

    that I is an ideal of A+I and that A ∩ I is an ideal of A.

Solutions

  1. The homomorphism preserves addition for every k but not multiplication. Consider the image of zz, φ(z2) = kz2. For the relation φ(z2) = φ(z)φ(z) = k2z2 to be satisfied, kz2 = k2z2 which is true only for 0 and 1.

  2. We notice first that it is an additive subgroup. In addition the product of a matrix with last row null by a generic matrix is again a matrix with null last row.

  3. It is not an additive subgroup: the sum of a matrix with null determinant can be a non singular matrix (1 0; 0 0) + (0 0; 0 1) = (1 0; 0 1).

  4. Only T is an ideal.

  5. Suppose the ideal I of 𝕂 is non-empty and the non-zero x ∈ I. Then since 𝕂 is a field it has inverse x−1. For yx−1 ∈ 𝕂 we have

    (yx−1) x = y

    so y ∈ I too and I = 𝕂.

  6. A(X) is an additive subgroup because if r1,r2 ∈ A(x) ⇒ r1x = 0 and r2x = 0 ⇒ (r1 − r2)x = 0 ⇒ r1 − r2 ∈ A(X). B(x) is a subgroup as well. We have that ∀ a ∈ A(X), ∀ r ∈ R, (ra)x = r(ax) = 0 hence ra ∈ A(X). Thus A(X) is a left ideal. The same resoning proves that B(X) is a right ideal. If X is left ideal, the ∀a ∈ A, r ∈ R, (ar)x = a(rx) with rx ∈ X = 0, thus A(X) is a two-sided ideal.

  7. A(X) = {(0,0; 0,0)} is a two-sided ideal. To find B(X) do the matrix product [x,0; y,0] [a,b; c,d] = [ax, xb; ay, yb] = 0 when a = b = 0, so B(X) = {[0,0; c,d]}. It is a right ideal because [0,0; c,d] [x,y; w,z] is again a matrix in B(x) but [x,y; w,z] [0,0; c,d] is not a matrix in B(X).

  8. Suppose that xm = 0 and yn = 0 and let r be any integer greater than m + n. By commutativity of R, we have the binomial theorem (same proof as usual) so in the expansion for (x + y)r every monomial has the form aij xiyj with i + j = r so either i > m or j > n or both. Notie that if

    xm = 0   ⇒   x(m+1) = x·xm = x·0 = 0

    Commutativity is required to express the product x·y·x·y·x·y·x··· as xk y(r-k). If the ring is not abelian the above expansion is not possible. If you expand (x+y)·(x+y)···(x+y) (r times), you have a sum of 2r terms, where each term is a product of r variables which are each either x or y.

    Thus, every monomial term is zero and hence x + y is nilpotent. Again by commutativity, (rx)m = rm xm = 0 so rx is nilpotent.

    Finally if you have commutativity you can write (a·x)n = an·xn and a·x = x·a, so (a·x)n = an·xn = (x·a)n. Thus N is a two-sided ideal.

  9. The only value is a0 = 0.

  10. It is an additive subgroup since [g(x)+f(x)](x2 + 5)] is also in I and it is a two-sided ideal because k(x)[f(x)(x2 + 5)] = [f(x)(x2 + 5)]k(x) for every k(x) ∈ ℚ[x]. Hence I is a two-sided ideal.

    By the division algorithm, for every p(x) ∈ ℚ[x] there exists q(x),r(x) ∈ ℚ[x] such that

    p(x) = (x2 + 3)q(x) + r(x)

    and that deg r(x) < deg(x2 + 3) = 2 whenever r(x) ≠ 0; put in another way: for every p(x) ∈ Q[x] there exists q(x) ∈ ℚ[x] and a,b ∈ ℚ such that p(x) = (x2 + 3)q(x) + (a + bx), which is the same as saying

    p(x) + I = (a+bx) + I.

    Thus ℚ[x]/I = {(a+bx) + I | a,b ∈ ℚ}.

  11. We must show a = 0 **in R/N** (ie a ~ 0), not a = 0 in R.

    (a+N)n = an + na(n−1)N + … + Nn

    so (a+N)n − an = na(n−1) N + … + Nn and the right side is nilpotent, so (a+N)n ~ an.

    Now we must show that an nilpotent ⇐⇒ a nilpotent: ak ∈ N, so there is some k' such that (ak)k' = 0, so akk' = 0, so a ∈ N, so a ~ 0.

  12. For a in I, b in J, a + b in I+J, we have (a+b)r = ar + br, which since ar is in I and br is in J is in I+J as well.

    For a in I, b in J, (∑ni=0 aibi)r = ∑ni=0 aibir, we have abr = a(br) and br is in J because J is an ideal, and a is in I ⇒ a(br) is in IJ.

  13. IJ ⊆ I ∩ J: A general element x of IJ is of the form x = i1 j1 +...+ in jn with either ik ∈ I, each jk in J. Since I is an ideal, each of those summands ik jk is also in I, so the sum x is also in I. Since J is an ideal, each ik jk ∈ J, so the sum x is also in J. So we get x ∈ I ∩ J.

    If 1 ∈ R, R(I ∩ J) = I ∩ J because R(I ∩ J) = {r⋅k | r ∈ R, k ∈ I ∩ J}, if 1 wasnt in R then R would only be a subset of I ∩ J. Hence I ∩ J = R(I ∩ J) = (I + J) (I ∩ J) = (II ∩ IJ) + (JI ∩ JJ) ⊆ IJ + JI ⊆ IJ.

    Alternatevely: Since I + J = R, there exist i in I, j in J with i + j = 1. So now take any x in I \cap J. So x = x1 = x(i + j) = ix + xj and now you see ix is in IJ since x is in J, and xj is in IJ since x is in I as well, so their sum is in IJ.

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