Homework 1: This contains exercises from Chapter 1 of Atiyah-Mcdonald

Exercise 1: Let \(x\) be a nilpotent element of a ring \(A\). Show that \(1+x\) is a unit of \(A\). Deduce that the sum of a nilpotent element and a unit is a unit.

Solution: The first part follows by observing that \((1+x)(1-x+x^2-x^3+x^4- \ldots \pm x^{n-1}) = 1 + x - x + x^2 - x^2 + \ldots +x^{n-1} - x^{n-1} \pm x^n =1\). Now to see this more generally, let \(y \in A\) be a unit. Because of this, we can write \((x+y) = (1+xy^{-1})\). Our previous result indicates that this should be similar. Observe that \((1+xy^{-1})(\sum _{k=0} ^{n-1} (-1)^k (xy^{-1})^k) = (1+xy^{-1})(1-xy^{-1}+(xy^{-1})^2 - \ldots \pm (xy^{-1})^{n-1}) = (1 +xy^{-1} - xy^{-1} - (xy^{-1})^2 + (xy^{-1})^2 - \ldots \pm (xy^{-1})^n) = 1\)

Exercise 12: A local ring, \(A\), contains no idempotent \(\neq 0, 1\).

Solution: Assume for sake of contradiction, \(A\) contains an idempotent element, \(x\), not equal to \(0\) or \(1\). Let \(m\) be the maximal ideal of \(A\). If we consider the residue of \(x\) over \(A/m\). We see that \( x+m = x^2 + m\), i.e. \(\begin{equation} 0 + m = (x^2 -x) +m = x(x-1) + m \end{equation}\). However, since \(A/m\) is a field, we can't have that \(x\) and \(x-1\) are zero divisors. This equality then forces \(x\) to be \(0\) or \(1\).

Exercise 15: Let \(A\) be a ring and let \(X\) be 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 \(\mathcal{a}\) is the ideal generated by \(E\), then \(V(E) = V(\mathcal{a}) = V(r(\mathcal{a}))\) (2): \(V(0) = X, V(1) = \emptyset\) (3): if \((E_i)_{i \in I}\) is any family of subsets of \(A\), then \(V(\cup _{i \in I} E_i) = \cap _{i \in I} V(E_i)\) (4): \(V(\mathcal{a} \cap \mathcal{b}) = V(\mathcal{ab}) = V(\mathcal{a}) \cup V(\mathcal{b})\) for any ideals, \(\mathcal{a}, \mathcal{b} \) of \(A\).

Solution:

(1): We show \(V(E) \subseteq V(a)\). Let \(I\) be a prime ideal containing \(E\) and \(x\) an element of \(\mathcal{a}\). So then \(x\) is of the form \(a_1e_1 + \ldots +a_ne_n\), where \(a_i \in A\) and \(e_i \in E\), \(n \in \mathbb{N}\). But notice that each individual \(a_i e_i\) is contained in \(I\) since it contains \(E\). So any element of \(\mathcal{a}\) is contained in \(I\). Hence, \(I \in V(\mathcal{a})\).

Now we show \(V(\mathcal{a}) \subseteq V(r(\mathcal{a}))\). Let \(I\) be a prime ideal containing \(\mathcal{a}\). Then for any \(x \in r(\mathcal{a})\), \(x^n \in I\) for some \(n >0\). However, since \(I\) is a prime ideal, we get that \(x \in I\). Thus, every element of \(r(\mathcal{a})\) is contained in \(I\), so \(I \in V(r(\mathcal{a}))\).

Now we show \(V(r(\mathcal{a})) \subseteq V(E)\). Let \(I\) be a prime ideal containing \(r(\mathcal{a})\). Since for any \(e \in E\), \(e^1 \in \mathcal{a}\), notice that \(E \subseteq r(\mathcal{a})\). So any ideal of \(r(\mathcal{a})\) contains \(E\) and so \(I \in V(E)\).

Thus we can conclude \(V(E) = V(\mathcal{a}) = V(r(\mathcal{a}))\).

(2): From the definition, \(V(0) = \{ I | I \textrm{ is prime }, 0 \in I \}\). Every ideal contains \(0\) as an element. So \(V(0)= X\), the set of all prime ideals of \(A\).

Similarly, \(V(1) = \{ I | I \textrm{ is prime }, 1 \in I \}\). If \(1\) is contained in any ideal, it becomes the whole ring. Our definition of ideal does not allow the whole ring as an ideal. So there is no prime ideal containing \(1\), hence \(V(1) = \emptyset\).

(3): Showing inclusion on the right. Let \(I\) be an ideal of \(V(\cup E_i)\). We wish to show \(I\) is an ideal for each \(E_i\). Since \(\cup E_i \subseteq I\), for any \(e_i \in E_i\), it follows that \(e \in I\). So \(E_i \subseteq I\). Hence, \(I \in \cap V(E_i)\)

Showing inclusion on the left: Let \(I\) be an ideal of \(\cap V(E_i)\). If \(e \in \cup E_i\), then \(e \in E_j\) for some \(j\) and since \(E_j \subseteq I\), it follows that \(e \in I\). So then \(I\) is an ideal containing \(\cup E_i\). Thus, \(I \in V(\cup E_i)\).

So we conclusde that \(\cap V(E_i) = V(\cup E_i)\)

(4): We first show that \(V(\mathcal{ab}) \subseteq V(\mathcal{a} \cap \mathcal{b})\). Let \(I \in V(\mathcal{ab}), x \in \mathcal{a} \cap \mathcal{b}\). Then \(x^2 \in I\) implies that \(x \in I\). So any ideal of \(V(\mathcal{ab})\) contains \(x \in \mathcal{a} \cap \mathcal{b}\). So then \(I \in V(\mathcal{a} \cap \mathcal{b})\).

Next we show that \(V(\mathcal{a} \cap \mathcal{b}) \subseteq V(\mathcal{a}) \cup V(\mathcal{b})\). Let \(I\) be an ideal containing \(\mathcal{a} \cap \mathcal{b}\). Then \(\mathcal{ab} \subset I\). Since \(I\) is prime, one of \(\mathcal{a}\) or \(\mathcal{b}\) is contained in \(I\). Thus, \(I \in V(\mathcal{a}) \cup V(\mathcal{b})\).

Finally, we show \(V(\mathcal{a}) \cup V(\mathcal{b}) \subseteq V(\mathcal{ab})\). If \(I\) is a prime ideal containing either \(\mathcal{a}\) or \(\mathcal{b}\). Then, \(I\) contains \(\mathcal{a} \cap \mathcal{b}\), which contains \(\mathcal{ab}\) as a subset. So \(I \in V(\mathcal{ab})\).