Topics
Ring homomorphisms, ideals, quotients, zero divisors, the isomorphism theorems, the Chinese remainder theorem
Irreducible and prime elements, nilpotent elements, units
Radicals, the nilradical, \(\spec R\) and \(\mspec R\)
Classes of rings: integral domains, Euclidean domains \(\implies\) PIDs \(\implies\) UFDs \(\implies\) integral domains, Dedekind domains, Noetherian and Artinian rings
Existence arguments using Zorn’s lemma
Localization
Ideals of $\mathbb Z$ containing relatively prime integers (Tag P-AMH-ALG-SG16-17)
(February 1982) Prove that if an ideal \(I\) of the ring \(\mathbb{Z}\) contains two relatively prime integers, then \(I=\mathbb{Z}\).
Solution
Proof. By hypothesis, there exist a,b∈I with gcd(a,b ) = 1. Thus, there are integers m,n∈ Z such thatma +nb = 1. But ma,nb∈I by the sticky property, so 1 = ma +nb∈I sinceI is closed under +. For any x∈ Z, then, we have x =x· 1∈I by the sticky property. So I = Z QED
Annihilators are ideals (Tag P-AMH-ALG-SG16-19)
(January 2010) Let \(R\) be a commutative ring and \(S\subseteq R\) a subset of \(R\). Define the annihilator of \(S\) in \(R\) to be \[ \operatorname{Ann}(S)=\{r\in R\mid rs=0\text{ for every }s\in S\}. \] Prove that \(\operatorname{Ann}(S)\) is an ideal of \(R\).
Solution
Proof. (Nonempty) We have 0∈ Ann(S), since 0s = 0∀s∈S. So Ann(S)⁄= ∅. (Closed under−) Given a,b ∈ Ann(S) and s∈ S, we have (a−b)s = as−bs = 0− 0 = 0, so that a−b∈ Ann(S). (Sticky) Given r∈ R and x∈ Ann(S), and given any s∈ S, we have ( rx)s = r(xs) = r(0) = 0. Thus,rx∈ Ann(S). Since R is commutative, xr =rx∈ Ann(S). QED
Quotients by ideals containing all commutators are commutative (Tag P-AMH-ALG-SG16-21)
(January 2009) Let \(I\subseteq R\) be an ideal of \(R\), and suppose that \(xy-yx\in I\) for every \(x,y\in R\). Prove that the quotient ring \(R/I\) is commutative.
Solution
Proof. GivenI +a,I +b∈R/I, our assumption on I implies that ab−ba∈I. By the coset relation, we obtain I +ab =I +ba. Then (I +a)(I +b) =I +ab =I +ba = (I +b)(I +a), so R/I is commutative, as desired QED
Reduced quotient rings and radical ideals (Tag P-AMH-ALG-SG16-22)
(January 2014) A nonzero element \(a\) of a ring is said to be nilpotent if there is a positive integer \(n\ge 1\) such that \(a^n=0\). (The element \(0\) itself is not said to be nilpotent.) Let \(R\) be a commutative ring, and let \(I\subseteq R\) be an ideal. Prove that the following two statements are equivalent. (a) The quotient ring \(R/I\) contains no nilpotents. (b) For every element \(b\in R\) such that \(b^m\in I\) for some positive integer \(m\ge 1\), we have \(b\in I\).
Solution
Proof. (=⇒) Given b∈R and m≥ 1 with bm∈I, we have (I +b)m =I +bm =I + 0. SinceI + 0 is the zero element of R/I, it follows that I +b is either nilpotent or zero in R/I. But by assumption (a), I +b is not nilpotent. Therefore, I +b =I + 0, which means b =b− 0∈I. (⇐=) SupposeI +b∈R/I is a nilpotent. Then there is some integerm≥ 1 such that (I +b)m =I +0. That is, I +bm = (I +b)m =I + 0, and hence bm =bm− 0∈I. By property (b), we have b∈I. Thus, I +b =I + 0, contradicting the assumption that I +b is nilpotent, and proving that R/I has no nilpotents. QED
A commutative unital ring is a field iff its only ideals are $\{0\}$ and $R$ (Tag P-3PTP5)
Prove that a commutative ring with unit is a field if and only if its only ideals are {0} and the whole ring
Solution
<1>1. If \(R\) is a field, then its only ideals are \(\{0\}\) and \(R\).
Let \(I\subseteq R\) be a nonzero ideal and choose \(0\ne a\in I\). Since \(R\) is a field, \(a^{-1}\in R\), so \[ 1=a^{-1}a\in I. \] Hence \(I=R\). Thus the only ideals are \(0\) and \(R\).
<1>2. Conversely, suppose the only ideals of the commutative unital ring \(R\) are \(\{0\}\) and \(R\). Then every nonzero element is a unit.
Take \(0\ne a\in R\). The principal ideal \((a)\) is nonzero, so by hypothesis \[ (a)=R. \] Hence \(1\in(a)\), so there exists \(b\in R\) with \[ ab=1. \] Thus \(a\) is a unit.
<1>3. Therefore \(R\) is a field.
By <1>2, every nonzero element of \(R\) is invertible. Since \(R\) is commutative with \(1\), this is exactly the definition of a field.
A polynomial of degree $2$ or $3$ in $k[x]$ is irreducible iff it has no root in $k$ (Tag P-HWBXR)
Let \(k\) be a field and let \(f(x) \in k[x]\) be a polynomial of degree \(\deg(f) \in \{2, 3\}\). Prove that \(f(x)\) is irreducible in \(k[x]\) if and only if \(f(x)\) has no roots in \(k\).
Solution
If \(f\) is irreducible and had a root \(a\in k\), then the factor theorem would give \[ f(x)=(x-a)g(x) \] with \(\deg g=\deg f-1\ge1\), a nontrivial factorization. Hence an irreducible polynomial of degree \(2\) or \(3\) has no root in \(k\).
Conversely, suppose \(f\) is reducible. Then \[ f=gh \] with \(\deg g,\deg h\ge1\) and \[ \deg g+\deg h=\deg f\in\{2,3\}. \] Therefore at least one factor has degree \(1\). Write that factor as \(ax+b\) with \(a\ne0\). It vanishes at \(-b/a\in k\), so \(f\) has a root in \(k\).
Thus, for degrees \(2\) and \(3\), \[ f\text{ irreducible}\quad\Longleftrightarrow\quad f\text{ has no root in }k. \]