## Ideals

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:

- $A$ has exactly one prime ideal.
- Every element of $A$ is either unit or nilpotent.
- $A/\mathfrak{N}$ 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).

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 $A$:

- $2x = 0$, for all $x \in A$.
- Every prime ideal $\mathfrak{p}$ is maximal, and $A/\mathfrak{p}$ is a field with two elements.
- 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:

- If $I$ is the ideal generated by $E$, then $V(E) = V(I) = V(\sqrt{I})$.
- $V(0) = X, V(1) = \phi$.
- If $(E_q)_{q \in Q}$ is the family of subsets of $A$ then: $$V \left(\bigcup_{q \in Q} E_q \right)= \bigcap_{q \in Q} V(E_q)$$
- $V(I \cap J) = V(IJ) = V(I) \cup V(J)$, 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 $\text{Spec}(A)$.

Proof:

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

## Modules

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,