[ac/hw] Homework


Prove that a variety \(X\) is irreducible if \(I(X)\) is a prime ideal.

The definition of an irreducible variety is as follows. Let \(X_1\) and \(X_2\) be two varieties. \[X = X_1 \cup X_2 \implies X = X_1 \vee X = X_2\]

Proof: Let us proceed using proof by contradiction, and assume that \(I\) is not prime. Let \(f\) and \(g\) be two prime ideals; i.e. \(fg \in I \implies f, g \in A \backslash I\), for ring \(A\). Now, define: \[J_1 = (I, f), J_2 = (I, g)\]

Then, \[V(J_1) \subsetneq X, V(J_2) \subsetneq X\]

which yields \(X = V_1 \cup V_2\), a reducible, as required \(\Box\)

If \(x\) is nilpotent \(\iff 1 - x\) is unit.

Proof: \((1 - x)(1 + x + \ldots + x^{n - 1}) = 1\); by binomial expansion, \(x^n = 0\) \(\Box\)

Prove that \(f = a_0 + \ldots + a_n x^n \in A[x]\) is unit iff \(a_1, \ldots, a_n\) are nilpotent, and \(a_0\) is unit.

Proof: Let \(fg = 1\), where \(g = \sum_m b_i x^i\). Clearly, \(a_0, b_0\) are units, and by collecting terms, we get:

\[a_n b_m = 0 \\ a_{n - 1} b_m + a_n b_{m - 1} = 0 \implies {a_n}^2 b_{m - 1} = 0 \\ \ldots \implies {a_n}^{m - 1} b_0 = 0\]

Since \(b_0 = 1\), it follows that \({a_n}^{m - 1} = 0\), which means \(a_n\) is nilpotent. By induction, we can prove that all \(a_i\)s for \(i > 0\) are nilpotent \(\Box\)

Prove that, in \(A[x]\), the Jacobson radical is equal to the nilradical.

Proof: Let \(f = a_0 + \ldots + a_n x^n \in \mathfrak{R}\). Then, \(1 + xf\) is unit, and by the previous problem, \(a_1, \ldots, a_n\) are all nilpotent, implying \(\mathfrak{R} = \mathfrak{N}\) \(\Box\)

Let \(A\) be a ring, and \(\mathfrak{N}\) be its nilradical. Then, show that the following are equivalent:

  1. \(A\) has exactly one prime ideal.
  2. Every element of \(A\) is either unit or nilpotent.
  3. \(A/\mathfrak{N}\) is a field.

Proof: (i) implies that there is exactly one maximal ideal, since every ideal must be contained within some maximal ideal: then, let \(\mathfrak{p} = \mathfrak{m}\) be this prime/maximal ideal. (iii) follows immediately, since, \(A/\mathfrak{m}\) is a field (see [ra/hw](/ra/hw)).

It follows from (i) that \(A\) is a local ring with maximal ideal \(\mathfrak{m}\). Let \(x \in A - \mathfrak{m}\). The ideal generated by \(x\) and \(\mathfrak{m}\) is \(A\), since \(\mathfrak{m}\) is maximal. Let \(y \in A, t \in \mathfrak{m}\), then \(xy + t = 1 \implies xy = 1 - t \in 1 + \mathfrak{m}\), and hence \(xy = 1 \implies y = 1/x\). This implies (ii), since every element \(y \in A\) is invertible.

Starting from (iii), we know that the only ideals of \(A/\mathfrak{N}\) are \((0)\) and \((1)\), by definition of a field. Hence, elements \(x \in A\) are of the form \(x + (0)\) and \(x + (1)\); the second coset generates the entire ring, while the first coset generates the single ideal in \(A\), and (i) follows from this \(\Box\)

Let ring \(A\) be such that every ideal not contained in the nilradical contains a nonzero idempotent (i.e. \(e^2 = e \neq 0\)). Prove that \(\mathfrak{N} = \mathfrak{R}\).

Proof: \(e \in \mathfrak{N} \iff 1 - ey\) is unit, for some \(y \in A\), but \(e \notin \mathfrak{R}\). Inverting logic, we get \(e \notin \mathfrak{N} \iff 1 - ey\) is not unit. Now, plugging \(y = e = e^2 \neq 0\), we get \(1 - e\) is not unit \(\implies 1 - e\) must be contained in some maximal ideal \(\implies e\) must be contained in every maximal ideal, and we arrive at our contradiction \(\Box\)

Let \(I\) be an ideal in ring \(A\) that is \(\neq (1)\). Then, show that \(I = \sqrt{I} \iff I\) is an intersection of prime ideals.

Proof: \(\sqrt{I} = \{x \in A \mid \exists n \gt 0 : x^n \in I\}\), by definition. Now suppose, for \(p \in A\), \(x^n = p \neq 0\). Then, \(px \in I\), by definition of an ideal, and the chain would not terminate at \(n\). Hence, \(\forall x \in I, x^n = 0\), and this is exactly the definition of the set of nilpotent elements.

Proof, the other way: \(I\) is the set of nilpotent elements. Let \(x \in I\) be any element; then, \(\exists n : x^n = 0 \in I\). Since every element of \(I\) is of this form, \(I = \sqrt{I}\) \(\Box\)

Prove that a local ring contains no idempotent \(\neq 0, 1\).

Proof: A local ring is defined as having exactly one maximal ideal, \(\mathfrak{m}\). Consider idempotent \(e \in \mathfrak{m} \implies 1 - e^2\) is unit, and is not contained within \(\mathfrak{m}\). Since \(e\) is idempotent, \(1 - e\) is unit as well. Now, since \(\mathfrak{m}\) and \(1 - e\) must generate \((1)\), \(e(1 - e) = 1 \implies e = 0 \vee e = 1\) \(\Box\)

A ring is boolean if \(x^2 = x\) for all \(x \in A\). Show that in a Boolean ring \(A\):

  1. \(2x = 0\), for all \(x \in A\).
  2. Every prime ideal \(\mathfrak{p}\) is maximal, and \(A/\mathfrak{p}\) is a field with two elements.
  3. Every finitely generated ideal in \(A\) is principal.

Proof: To prove (ii), let \(x + P \in A/\mathfrak{p}, x \notin \mathfrak{p}\). Then, \(x(x - 1) = 0 \in \mathfrak{p}\). Since \(\mathfrak{p}\) is prime, and \(x \notin \mathfrak{p}\), \(x - 1 \in P\) implying that \(x + \mathfrak{p}\) is invertible, and hence \(A/\mathfrak{p}\) is a field.

Let \(A\) be a ring, and \(X\) the set of all prime ideals of \(A\). For each subset \(E\) of \(A\), let \(V(E)\) denote the set of all prime ideals of \(A\) which contain \(E\). Prove that:

  1. If \(I\) is the ideal generated by \(E\), then \(V(E) = V(I) = V(\sqrt{I})\).
  2. \(V(0) = X, V(1) = \phi\).
  3. If \((E_q)_{q \in Q}\) is the family of subsets of \(A\) then: \[V \left(\bigcup_{q \in Q} E_q \right)= \bigcap_{q \in Q} V(E_q)\]
  4. \(V(I \cap J) = V(IJ) = V(I) \cup V(J)\), for ideals \(I, J\).

These results show that \(V(E)\) satisfy the axioms for a closed set in a topological space. The resulting topology is called a `Zariski topology`, and \(X\) is called the `prime spectrum` of \(A\), written \(\text{Spec}(A)\).


Prove that \(\text{Spec } A\) is irreducible iff \(\mathfrak{N}\) of \(A\) is a prime ideal.


Prove that \((\mathbb{Z}/m\mathbb{Z}) \otimes_Z (\mathbb{Z}/n\mathbb{Z}) = 0\) if \(m, n\) are coprime.

Let \(A\) be a ring, \(I\) and ideal, and \(M\) a module. Prove that \((A/I) \otimes_A M\) is isomorphic to \(M/IM\).

Proof: Tensor exact sequence \(0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0\) with \(M\).

Let \(M, N\) be finitely-generated \(A\)-modules. If \(M \otimes N = 0\) show that \(M = 0\) or \(N = 0\).

Proof: Let \(\mathfrak{m}\) be a maximal ideal, and \(k = A/\mathfrak{m}\) the reside field. Let \(M_k = k \otimes_A M \cong M/\mathfrak{m} M\). By Nakayama's lemma, \(M_k = 0 \implies M = 0\). But \(M \otimes_A N = 0 \implies (M \otimes_A N)_k = 0 \implies M_k \otimes_k N_k = 0\), which in turn implies \(M_k = 0\) or \(N_k = 0\) since \(M_k, N_k\) are vector spaces over a field.

Let \(M_i\) be a family of \(A\)-modules, and \(M\) be their direct sum. Then, prove \(M\) is flat \(\iff\) each \(M_i\) is flat.

If \(\mathfrak{p}\) is a prime ideal in \(A\), prove that \(\mathfrak{p}[x]\) is a prime ideal in \(A[x]\).

Rings of fractions

If \(S\) is a multiplicatively closed set of ring \(A\), and \(M\) a finitely-generated \(A\)-module, prove that \(S^{-1} A = 0\) iff there exists \(s \in S\) such that \(sM = 0\).

Let \(I\) be an ideal of ring \(A\), and \(S = 1 + I\). Show that \(S^{-1} I\) is contained within the Jacobson radical of \(S^{-1} A\). Use this and Nakayama's lemma to show that there exists \(x \equiv 1 \text{ mod } I\) such that \(xM = 0\), when \(IM = M\).

Proof: To prove the second part, notice that if \(M = IM\), then \(S^{-1} M = (S^{-1} I)(S^{-1} M)\), hence by Nakayama's lemma, we have \(S^{-1} M = 0\). Now, from the previous exercise,