Let rings be commutative with identity.
- Show that every proper ideal is contained in a maximal ideal.
- If \(R\) is a PID and \(x\in R\) is a nonzero nonunit, show that \(x\) is irreducible iff \((x)\) is maximal.
- Show that intersections, products, and sums of ideals are ideals.
- Show that the union of two ideals need not be an ideal.
- Show that every nonzero ring has a maximal ideal.
- Show that \(I\) is maximal iff \(R/I\) is a field.
- Show that \(I\) is prime iff \(R/I\) is an integral domain.
- Show that \[ \bigcup_{\mathfrak m\in\operatorname{MaxSpec}R}\mathfrak m =R\setminus R^\times. \]
- Show that \(\operatorname{MaxSpec}R\subseteq\operatorname{Spec}R\), and give an example of strict containment and one of equality.
- Show that every prime ideal is radical.
- Show that the nilradical is \(\sqrt{(0)}\).
- Show that \[ \sqrt{IJ}=\sqrt I\cap\sqrt J. \]
- If \(R\) is an integral domain and \(\operatorname{Spec}R\subseteq\operatorname{MaxSpec}R\), show that \(R\) is a UFD.
- Show that a Noetherian ring has every ideal finitely generated.
Solution
1. Let \(I\subsetneq R\). Consider the set of proper ideals containing \(I\), ordered by inclusion. The union of a chain is again an ideal, and it is proper: if its union contained \(1\), then one member of the chain would contain \(1\). By Zorn’s lemma there is a maximal member, i.e. a maximal ideal containing \(I\).
2. Suppose \(x\) is irreducible and\[ (x)\subseteq(a)\subseteq R. \]Then \(x=ar\). Irreducibility gives either \(a\) a unit, so \((a)=R\), or \(r\) a unit, so \((a)=(x)\). Hence \((x)\) is maximal. Conversely, if \((x)\) is maximal and \(x=ab\), then\[ (x)\subseteq(a)\subseteq R. \]Maximality forces \((a)=(x)\) or \((a)=R\), so one of \(a,b\) is a unit. Thus \(x\) is irreducible.
3. Arbitrary intersections of ideals are ideals. For ideals \(I,J\),\[ I+J=\{a+b:a\in I,b\in J\} \]is an ideal, and\[ IJ=\left\{\sum_{k=1}^r a_kb_k:a_k\in I,b_k\in J\right\} \]is an ideal by closure under addition and multiplication by arbitrary elements of \(R\).
4. In \(\ZZ\), the ideals \((2)\) and \((3)\) have union that is not an ideal: \(2,3\) lie in the union but\[ 2+3=5 \]does not.
5. If \(R\ne0\), then \((0)\) is proper. By 1 it lies in a maximal ideal.
6. Ideals of \(R/I\) correspond to ideals of \(R\) containing \(I\). Thus \(I\) is maximal exactly when \(R/I\) has no ideals except \(0\) and itself. A nonzero commutative ring with identity has this property exactly when it is a field.
7. One has\[ ab\in I\iff (a+I)(b+I)=0\text{ in }R/I. \]Thus \(I\) is prime exactly when \(R/I\) has no zero divisors, i.e. is an integral domain.
8. A unit cannot lie in a proper ideal, so no unit lies in any maximal ideal. Conversely, if \(x\) is a nonunit, then \((x)\) is proper, hence by 1 lies in some maximal ideal. Therefore the union of maximal ideals is exactly the set of nonunits.
9. Every maximal ideal is prime by <1>6–<1>7, since every field is a domain. Hence\[ \operatorname{MaxSpec}R\subseteq\operatorname{Spec}R. \]For \(R=\ZZ\) the containment is strict because \((0)\) is prime but not maximal. For a field \(k\),\[ \operatorname{Spec}k=\operatorname{MaxSpec}k=\{(0)\}. \]
10. Let \(\mathfrak p\) be prime and suppose \(x^n\in\mathfrak p\). Repeated primality applied to\[ x^n=x\cdot x^{n-1} \]shows \(x\in\mathfrak p\). Thus \(\mathfrak p=\sqrt{\mathfrak p}\).
11. By definition,\[ \sqrt{(0)}=\{x\in R:x^n=0\text{ for some }n\ge1\}, \]which is exactly the nilradical.
12. Since \(IJ\subseteq I\cap J\),\[ \sqrt{IJ}\subseteq\sqrt I\cap\sqrt J. \]Conversely, if \(x\in\sqrt I\cap\sqrt J\), choose \(m,n\) with \(x^m\in I\) and \(x^n\in J\). Then\[ x^{m+n}=x^mx^n\in IJ, \]so \(x\in\sqrt{IJ}\).
13. Since \(R\) is a domain, \((0)\) is prime. The hypothesis therefore makes \((0)\) maximal. Hence\[ R/(0)=R \]is a field by
6. Every field is a UFD.
14. If “Noetherian” is defined by the ascending-chain condition, suppose an ideal \(I\) were not finitely generated. Choose \(a_1\in I\), then inductively choose\[ a_{n+1}\in I\setminus(a_1,\ldots,a_n). \]This produces a strictly ascending chain\[ (a_1)\subsetneq(a_1,a_2)\subsetneq(a_1,a_2,a_3)\subsetneq\cdots, \]contradicting the ACC. Hence every ideal is finitely generated.