The implications among the classes of rings, with a ring separating each class from the next, are on Which kind of ring is this?.
Elements
Let \(R\) be an integral domain. Elements \(a,b\in R\) are associates if \(a=ub\) for a unit \(u\), so factorizations are compared up to units and associates. A nonzero nonunit \(p\) is irreducible if \(p=ab\) implies that \(a\) or \(b\) is a unit, and prime if \(p\divides ab\) implies \(p\divides a\) or \(p\divides b\).
Divisibility of elements (Tag D-AVBIP)
Let \(R\) be a commutative ring and let \(q, r\in R\). The element \(r\) is divisible by \(q\), written \(q\divides r\), if there exists \(c \in R\) such that \(r = qc\).
Units (Tag D-QQIQZ)
Let \(R\) be a ring. An element \(r\in R\) is a unit if there exists \(s\in R\) such that \(rs = sr = 1\). The set of units of \(R\) is denoted \(R^{\times}\).
If \(r\in R^{\times}\), the element \(s\) with \(rs=sr=1\) is unique: if also \(rs'=s'r=1\), then \(s'=s'(rs)=(s'r)s=s\). It is written \(r^{-1}\), and \((R^{\times}, \cdot)\) is a group, abelian when \(R\) is commutative.
Irreducible element (Tag D-TO3IY)
Let \(R\) be a commutative ring. A nonzero element \(r\in R\) that is not a unit is irreducible if for all \(a,b\in R\), \[ r=ab \implies a \in R^{\times} \text{ or } b\in R^{\times}. \]
Irreducible element of a ring (Tag FD-6LLJF)
Let \(R\) be a commutative ring. An element \(r\in R\) is irreducible if \(r\) is nonzero, \(r\) is not a unit, and whenever \(r=ab\) with \(a,b\in R\), \(a\) or \(b\) is a unit.
Irreducible element of a ring (Tag FD-R4K7Z)
Let \(R\) be a commutative ring. An element \(p\in R\) is irreducible if \(p\) is nonzero, \(p\notin R^\times\), and whenever \(p=ab\) with \(a,b\in R\), \(a \in R^\times\) or \(b\in R^\times\).
Prime element (Tag D-AWSKI)
Let \(R\) be a commutative ring. A nonzero element \(p\in R\) that is not a unit is prime if for all \(a,b \in R\), \[ p\divides ab \implies p\divides a \text{ or } p\divides b, \] where \(\divides\) denotes divisibility.
Prime element of a ring (Tag FD-S52W7)
Let \(R\) be a commutative ring. An element \(r\in R\) is prime if \(r\) is nonzero, \(r\) is not a unit, and for all \(a,b\in R\), \(r\divides ab\) implies \(r\divides a\) or \(r\divides b\).
Associate elements (Tag D-R4H6F)
Let \(R\) be a commutative ring. Elements \(a, b\in R\) are associates if there exists a unit \(u\in R^{\times}\) such that \(a = ub\).
Let \(R\) be an integral domain and \(a,b\in R\). Then \(a\) and \(b\) are associates if and only if \(a\divides b\) and \(b\divides a\).
If \(a=ub\) with \(u\) a unit, then \(b\divides a\), and \(b=u^{-1}a\) gives \(a\divides b\). Conversely, suppose \(b=ac\) and \(a=bd\) with \(c,d\in R\), in the sense of divisibility. If \(a=0\), then \(b=0\) and \(a=1\cdot b\). If \(a\neq 0\), then \(a=acd\) gives \(a(1-cd)=0\), so \(cd=1\) because \(R\) is an integral domain; thus \(d\) is a unit and \(a=db\).
(Prime and irreducible elements)
In an integral domain every prime element is irreducible. In a unique factorization domain every irreducible element is prime. In \(\ZZ[\sqrt{-5}]\) the element \(3\) is irreducible and not prime, since \(3\) divides \(9=(2+\sqrt{-5})(2-\sqrt{-5})\) and divides neither factor.
Classes of rings
A commutative ring \(R\neq 0\) is an integral domain if it has no nonzero zero divisors, equivalently if \(ab=ac\) and \(a\neq0\) imply \(b=c\). It is a field if every nonzero element is a unit.
Division ring (Tag D-7O2CH)
A division ring, or skew field, is a ring \(R\), not necessarily commutative, with \(1\neq 0\) in which every nonzero element is a unit, that is, \(R\setminus\theset 0 \subseteq \unitsof{R}\).
Zero divisor (Tag D-4I3SL)
Let \(R\) be a ring. An element \(r\in R\) is a zero divisor if there exists \(a\in R\setminus\theset 0\) such that \(ar = ra = 0\).
If \(R\) is commutative, then \(r\in R\) is a zero divisor if and only if the \(R\)-linear map \[ R \to R, \qquad x \mapsto rx \] is not injective.
Zero divisor (Tag FD-S62UB)
Let \(R\) be a commutative ring. An element \(r\in R\) is a zero divisor if there exists a nonzero \(x\in R\) such that \(rx=0\).
Equivalently, the \(R\)-linear map \(R\to R\), \(x\mapsto rx\), is not injective, since its kernel is \(\theset{x\in R \st rx=0}\).
Integral domain (Tag D-QJ3QL)
A nonzero commutative ring \(R\) is an integral domain if it has no nonzero zero divisors: \[ \text{for all } a, b\in R,\quad ab = 0 \implies a = 0 \text{ or } b = 0. \]
Integral domain (Tag FD-2A5XH)
An integral domain is a nonzero commutative ring \(R\) with no nonzero zero divisors: for \(a,b\in R\), \(ab=0\) implies \(a=0\) or \(b=0\).
Field (Tag D-UI6CU)
A field is a commutative division ring: a commutative ring with \(1\neq 0\) in which every nonzero element is a unit.
Simple ring (Tag FD-SNOTW)
A nonzero ring \(R\) is simple if its only two-sided ideals are \((0)\) and \(R\).
$A$ is a field iff $A$ is a simple ring iff every homomorphism from $A$ to a nonzero field is injective (Tag E-HOJKE)
Show that TFAE:
\(A\in \Field\)
\(A\) is a simple ring, so \(\Id(A) = \ts{ 0, A }\).
If \(B\in \Field\) is nonzero then every ring morphism \(A\to B\) is injective.
Solution
Let \(A\) be a commutative ring with \(1\neq0\); ring homomorphisms preserve \(1\). Number the three conditions (1), (2), (3) in the order stated.
<1>1. (1) implies (2).
Let \(I\) be a nonzero ideal of the field \(A\) and \(0\neq x\in I\). Then \(1=x^{-1}x\in I\), so \(a=a\cdot1\in I\) for every \(a\in A\) and \(I=A\).
<1>2. (2) implies (3).
Let \(\varphi\colon A\to B\) be a ring homomorphism to a field. \(\ker\varphi\) is an ideal, and \(\varphi(1)=1\neq0\) gives \(\ker\varphi\neq A\). By (2), \(\ker\varphi=0\), so \(\varphi\) is injective.
<1>3. (3) implies (1).
Let \(0\neq x\in A\) and suppose \((x)\neq A\). The proper ideal \((x)\) lies in a maximal ideal \(\mathfrak m\) (Zorn’s lemma), and \(\pi\colon A\to A/\mathfrak m\) is a homomorphism to a field. By (3), \(\mathfrak m=\ker\pi=0\), so \(x\in\mathfrak m=0\), a contradiction. Hence \((x)=A\), so \(x\) is a unit, and \(A\) is a field.
<1>4. Q.E.D.
Steps <1>1, <1>2 and <1>3 give \((1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)\).
Euclidean domains, principal ideal domains, and unique factorization domains
For integral domains, \[ \text{Euclidean domain}\implies\text{PID}\implies\text{UFD}\implies\text{integral domain}, \] and none of these implications reverses. In a Euclidean domain the division algorithm computes greatest common divisors; in a principal ideal domain \(\gens{a,b}=\gens{d}\) for a greatest common divisor \(d\) of \(a\) and \(b\); and in a unique factorization domain every nonzero nonunit is a product of irreducible elements, unique up to order and associates.
Principal ideal (Tag D-D7VK2)
Let \(R\) be a commutative ring. An ideal \(I \normal R\) is principal if there exists \(a\in R\) such that \(I = \gens{a} = Ra\).
Principal ideal domain (Tag D-HTIL5)
An integral domain \(R\) is a principal ideal domain if every ideal of \(R\) is principal.
Prime ideals, factorization, and chain conditions in PIDs (Tag E-KB4GA)
Let \(R\) be a PID.
- Prove that every nonzero prime ideal is maximal.
- Prove that \(R\) is Noetherian.
- Prove that every nonzero nonunit factors into finitely many irreducibles.
- Prove that every nonzero maximal ideal is generated by an irreducible element.
- Prove that \(\mathbb Z\) is Noetherian but not Artinian.
Solution
<1>1. Every nonzero prime ideal is maximal.
Let \((p)\) be a nonzero prime ideal and suppose \[ (p)\subseteq(a)\subseteq R. \] Then \(p=ra\). Since \(p=ra\in(p)\) and \((p)\) is prime, either \(a\in(p)\) or \(r\in(p)\). In the first case \((a)=(p)\). In the second, \(r=sp\), so \[ p=spa. \] Since \(R\) is a domain and \(p\ne0\), \(sa=1\), hence \((a)=R\). Thus \((p)\) is maximal.
<1>2. Every PID is Noetherian.
Every ideal is principal, hence finitely generated. A ring in which every ideal is finitely generated is Noetherian.
<1>3. Every nonzero nonunit factors into finitely many irreducibles.
Suppose not. Choose a nonzero nonunit \(a_0\) with no factorization into irreducibles. Then \(a_0\) is reducible, say \[ a_0=b_1a_1 \] with both factors nonunits. At least one factor, say \(a_1\), also has no factorization into irreducibles. Then \[ (a_0)\subsetneq(a_1). \] Repeating produces an infinite strictly ascending chain of ideals, contradicting Noetherianity. Hence the process terminates and gives a finite irreducible factorization.
<1>4. Every nonzero maximal ideal is generated by an irreducible.
Write \(\mathfrak m=(m)\ne(0)\). Since \(\mathfrak m\) is proper, \(m\) is not a unit. If \(m=ab\), then \[ (m)\subseteq(a)\subseteq R. \] Maximality gives \((a)=(m)\) or \((a)=R\), so respectively \(b\) or \(a\) is a unit. Thus \(m\) is irreducible.
<1>5. \(\mathbb Z\) is Noetherian but not Artinian.
It is a PID, hence Noetherian. But \[ (2)\supsetneq(4)\supsetneq(8)\supsetneq\cdots \] is an infinite strictly descending chain of ideals, so \(\mathbb Z\) is not Artinian.
Unique factorization domain (Tag D-INULL)
An integral domain \(R\) is a unique factorization domain if every nonzero \(r\in R\) admits a decomposition \[ r = u \prod_{i=1}^n p_i \] with \(n\geq 0\), \(u\in R^{\times}\) a unit, and \(p_1,\ldots,p_n\) irreducible, and this decomposition is unique up to order and associates: if also \(r = v\prod_{j=1}^m q_j\) with \(v\in R^{\times}\) and each \(q_j\) irreducible, then \(m=n\) and there is a permutation \(\sigma\) of \(\theset{1,\ldots,n}\) such that \(p_i\) and \(q_{\sigma(i)}\) are associates for every \(i\).
Unique factorization domain (Tag FD-25SUQ)
An integral domain \(R\) is a unique factorization domain if every nonzero \(x\in R\) can be written as \[ x=u\prod_{i=1}^n p_i \] with \(n\geq 0\), \(u\in R^\times\) a unit, and \(p_1,\ldots,p_n\) irreducible, and this decomposition is unique up to units and order: if also \(x=v\prod_{j=1}^m q_j\) with \(v\in R^\times\) and each \(q_j\) irreducible, then \(m=n\) and there is a permutation \(\sigma\) of \(\theset{1,\ldots,n}\) such that \(p_i\) and \(q_{\sigma(i)}\) are associates for every \(i\).
Euclidean domain (Tag D-NKRGN)
An integral domain \(R\) is Euclidean if there exists a function \(d\colon R\to \ZZ_{\geq 0}\) such that for all \(x,y\in R\) with \(y\neq 0\) there exist \(q,r\in R\) with \(x = qy + r\) and either \(r=0\) or \(d(r) < d(y)\). Such a function \(d\) is a Euclidean function on \(R\).
Euclidean domain (Tag FD-GXXBV)
An integral domain \(R\) is a Euclidean domain if there exists a function \(f\colon R\setminus\theset{0} \to \ZZ^{\geq 0}\), a Euclidean function, such that for all \(a\in R\) and all nonzero \(b\in R\) there exist \(q, r\in R\) with \(a = bq + r\) and either \(r=0\) or \(f(r) < f(b)\).
Other classes of rings
A ring is Noetherian if every ascending chain of ideals stabilizes.
A ring is reduced if its only nilpotent element is \(0\).
A ring is local if it has a unique maximal ideal.
A valuation ring is an integral domain \(R\) with fraction field \(K\) such that \(x\in R\) or \(\inverseof{x}\in R\) for every \(x\in K^\times\); a discrete valuation ring is a valuation ring whose valuation has value group \(\ZZ\).
In a Dedekind domain every nonzero proper ideal is a product of prime ideals, uniquely up to order.
A Noetherian local ring \((R,\mfm)\) is regular if \(\dim_{R/\mfm}\mfm/\mfm^2=\dim R\).
Noetherian ring (Tag D-TZXBO)
A ring \(R\) is Noetherian if it satisfies the ascending chain condition on ideals: for every chain of ideals \(I_1 \subseteq I_2 \subseteq \cdots\) of \(R\) there exists \(N\geq 1\) such that \(I_N = I_{N+1} = \cdots\).
Reduced ring (Tag D-PQHHJ)
A ring \(R\) is reduced if \(R\) contains no nonzero nilpotent elements.
Reduced ring (Tag FF-U4FHF)
A commutative ring \(R\) is reduced if and only if it has no nonzero nilpotent elements, that is, its nilradical \(\nilrad{R}\) is \(0\).
Local ring (Tag D-TGB4R)
A commutative ring \(R\) is local if it has a unique maximal ideal \(\mfm\), that is, \(\mspec R = \theset{ \mfm }\). The residue field of the local ring \((R,\mfm)\) is \(\kappa \coloneqq R/\mfm\).
A ring in which every element is a unit or nilpotent is local (Tag E-YYL5U)
Show that if \(R\) is a nonzero ring where every element is either a unit or nilpotent, then \(R\) is local.
Solution
Let \(N(R)\) denote the nilradical, i.e. the set of nilpotent elements of \(R\).
<1>1. No nilpotent element of the nonzero ring \(R\) is a unit.
If \(a^m=0\) and \(a\) were a unit, then multiplying by \(a^{-m}\) would give \(1=0\), contradicting that \(R\) is nonzero.
<1>2. The set of nonunits of \(R\) is exactly \(N(R)\).
By hypothesis every element of \(R\) is either a unit or nilpotent. By <1>1 these possibilities are disjoint. Hence an element is a nonunit exactly when it is nilpotent.
<1>3. The set of nonunits is a proper ideal.
Because \(R\) is commutative, its nilpotent elements form the nilradical \(N(R)=\sqrt{(0)}\), which is an ideal. It is proper because \(1\) is not nilpotent in a nonzero ring. By <1>2, this ideal is exactly the set of nonunits.
<1>4. The ring \(R\) has a unique maximal ideal.
Since \(N(R)\) is proper, it is contained in some maximal ideal \(\mathfrak m\). Every element of a proper ideal is a nonunit, so \(\mathfrak m\subseteq N(R)\) by <1>2. Thus \(N(R)=\mathfrak m\).
If \(\mathfrak m'\) is any maximal ideal, then every element of \(\mathfrak m'\) is a nonunit, hence lies in \(N(R)=\mathfrak m\). Maximality gives \(\mathfrak m'=\mathfrak m\). Thus the maximal ideal is unique.
<1>5. Therefore \(R\) is local.
A nonzero commutative ring is local precisely when it has a unique maximal ideal. By <1>4, that ideal is \(N(R)\).
$R_p$ is a local ring for $p\in\operatorname{Spec} R$ (Tag E-K4SU4)
Show that if \(\mathfrak p\in\operatorname{Spec}(R)\), then \(R_\mathfrak p\) is local with unique maximal ideal \(\mathfrak pR_\mathfrak p\).
Solution
Let \(S=R\setminus\mathfrak p\), so \(R_\mathfrak p=S^{-1}R\). The extension \[ \mathfrak m=\mathfrak pR_\mathfrak p \] is a proper ideal.
We claim that every element outside \(\mathfrak m\) is a unit. Let \(r/s\in R_\mathfrak p\) with \(r/s\notin\mathfrak m\). If \(r\in\mathfrak p\), then \(r/s\in\mathfrak m\), so necessarily \(r\notin\mathfrak p\), hence \(r\in S\). Therefore \[ \frac{s}{r}\in R_\mathfrak p \] and \[ \frac rs\frac sr=1. \] Thus every nonunit lies in \(\mathfrak m\).
Any proper ideal contains no unit, so every proper ideal of \(R_\mathfrak p\) is contained in \(\mathfrak m\). Hence \(\mathfrak m\) is the unique maximal ideal, and \(R_\mathfrak p\) is local.
A ring is local if $R\setminus\mfm$ consists of units, or if $1+\mfm$ consists of units (Tag E-PQ3FR)
Suppose \(\mfm \in \mspec R\) is a proper maximal ideal. Show that under either of the following two conditions, \(R\) is local:
\(R\sm \mfm \subseteq \unitsof{R}\), so every element of \(R\sm \mfm\) is a unit, or
\(1 + \mfm \subseteq \unitsof{R}\)
Solution
Let \(R\) be a commutative ring with identity and \(N \da R\sm \unitsof{R}\) its set of nonunits.
<1>1. If \(R\sm \mfm \subseteq \unitsof{R}\), then \(\mfm\) is the only maximal ideal of \(R\).
The proper ideal \(\mfm\) contains no unit, so \(\mfm\subseteq N\), and the hypothesis gives \(N\subseteq\mfm\); hence \(\mfm=N\). A proper ideal \(I\) contains no unit, so \(I\subseteq N=\mfm\).
<1>2. If \(1 + \mfm \subseteq \unitsof{R}\), then \(R\sm \mfm \subseteq \unitsof{R}\), so \(R\) is local by step <1>1.
Let \(r\in R\sm \mfm\). By maximality, \(\mfm+\gens{r} = R\), so \(rt + m = 1\) for some \(t\in R\) and \(m\in \mfm\). Then \(rt = 1-m \in 1 + \mfm \subseteq \unitsof{R}\), so \(r\) is a unit.
Dedekind domain (Tag D-WUGPG)
A Dedekind domain is an integral domain \(R\) in which every nonzero ideal \(I\) factors as a product \(I=\mfp_1\cdots\mfp_n\) of nonzero prime ideals, with \(n\geq 0\) and the empty product equal to \(R\), and this factorization is unique up to the order of the factors.
A Dedekind domain is a PID if and only if it is a UFD (Tag E-M6J67)
Show that a Dedekind domain \(R\) is a PID iff \(R\) is a UFD.
Solution
<1>1. If \(R\) is a PID, then \(R\) is a UFD.
In a PID every nonzero nonunit factors into irreducibles, because the ascending chain condition on principal ideals holds, and every irreducible element is prime, because it generates a maximal ideal. Hence factorization is unique up to order and associates.
<1>2. Suppose now that \(R\) is a Dedekind domain and a UFD. Then every nonzero prime ideal of \(R\) is principal.
A Dedekind domain is Noetherian and every nonzero prime ideal has height \(1\). In a Noetherian UFD, every height-\(1\) prime ideal is principal. Thus for every nonzero prime ideal \(\mathfrak p\subset R\) there is a prime element \(\pi_{\mathfrak p}\in R\) with \[ \mathfrak p=(\pi_{\mathfrak p}). \]
<1>3. Every nonzero ideal of \(R\) is principal.
Let \(0\neq I\subseteq R\). Since \(R\) is Dedekind, \(I\) has a factorization \[ I=\mathfrak p_1^{e_1}\cdots \mathfrak p_r^{e_r} \] into nonzero prime ideals. By <1>2, write \[ \mathfrak p_i=(\pi_i). \] Then \[ I=(\pi_1)^{e_1}\cdots(\pi_r)^{e_r} =\bigl(\pi_1^{e_1}\cdots\pi_r^{e_r}\bigr), \] so \(I\) is principal. The zero ideal is \((0)\), hence is principal as well.
<1>4. Therefore a Dedekind domain is a PID if and only if it is a UFD.
The forward implication is <1>1. For the reverse implication, <1>3 shows that every ideal of \(R\) is principal, so \(R\) is a PID.
Valuation ring (Tag D-HWLVG)
An integral domain \(R\) with field of fractions \(K\) is a valuation ring if for every nonzero \(x\in K\), \(x\in R\) or \(x^{-1} \in R\).
Discrete valuation ring (Tag D-VK2KZ)
A discrete valuation ring, or DVR, is a principal ideal domain that is a local ring and is not a field.
Regular local rings, regular rings, and DVRs (Tag D-DEFREGLR)
A Noetherian local ring \((A,\mm)\) is a regular local ring if \[ \dim A = \dim_{A/\mm} \mm/\mm^2 . \] A Noetherian ring \(A\) is a regular ring if \(A_\mfp\) is a regular local ring for every prime \(\mfp\).
(valuation)
Let \(K\) be a field and \(G\) a totally ordered abelian group. A valuation of \(K\) with values in \(G\) is a group homomorphism \(v \colon K^\times \to G\) such that \(v(x+y) \geq \min(v(x), v(y))\) for all \(x, y \in K^\times\) with \(x + y \neq 0\). Its valuation ring is \(R = \ts{x \in K^\times \st v(x) \geq 0} \cup \ts{0}\). A domain \(R\) is a valuation ring if it is the valuation ring of some valuation of \(\operatorname{Frac} R\), and the valuation is discrete if \(G = \ZZ\). [@Har10a, §I.6]
By Krull’s height theorem, \(\dim A \leq \dim_{A/\mm}\mm/\mm^2\) for every Noetherian local ring, and \(\dim_{A/\mm}\mm/\mm^2\) is the minimal number of generators of \(\mm\). If \(A\) is regular, then \(\mm\) is generated by a regular sequence, and \(A\) is Cohen–Macaulay.
A regular local ring is a domain, is a UFD by the Auslander–Buchsbaum theorem, and is therefore integrally closed. By Serre’s theorem, a Noetherian local ring is regular if and only if its residue field has finite projective dimension, if and only if it has finite global dimension; consequently every localization of a regular local ring at a prime is regular.
For a Noetherian local domain of dimension one, the following are equivalent: it is regular; it is a DVR; it is integrally closed; \(\mm\) is principal. Hence on a Noetherian integral scheme regular in codimension one, the local ring at the generic point of each prime divisor \(Y\) is a DVR, and its valuation \(v_Y\) gives the coefficients of the principal Weil divisor \(\operatorname{div}(f)\) (D-5PQ5W).
Regular local rings and regular rings (Tag FF-CQSNC)
A Noetherian local ring \((R,\mfm)\) with residue field \(k=R/\mfm\) is regular if \(\mfm\) can be generated by \(\dim R\) elements, where \(\dim R\) is the Krull dimension of \(R\). By Nakayama's lemma for a local ring, the minimal number of generators of \(\mfm\) is \(\dim_k\mfm/\mfm^2\), so \(R\) is regular if and only if \(\dim_k\mfm/\mfm^2=\dim R\).
A commutative Noetherian ring \(R\) is regular if the localization \(R_{\mfp}\) is a regular local ring for every prime ideal \(\mfp\subseteq R\).
By Krull’s height theorem, the extension of Krull's Hauptidealsatz to ideals generated by \(r\) elements, \(\dim R\le\dim_k\mfm/\mfm^2\) for every Noetherian local ring \((R,\mfm)\); regularity is the case of equality.
Structure theorems
An \(R\)-module is simple if it is nonzero and has no submodules other than \(0\) and itself, and semisimple if it is a direct sum of simple modules. By the Artin–Wedderburn theorem, a ring that is semisimple as a left module over itself is isomorphic to a finite product \(\prod_i M_{n_i}(D_i)\) of matrix rings over division rings \(D_i\). By Wedderburn’s little theorem, every finite division ring is a field.
Simple modules (Tag D-4KM4P)
Let \(R\) be a ring. An \(R\)-module \(M\) is simple if \(M\neq 0\) and every submodule \(M' \leq M\) is either \(0\) or \(M\). A commutative ring \(R\) is simple if it is simple as an \(R\)-module, that is, if \(R\neq 0\) and its only ideals are \(0\) and \(R\).
Semisimple modules (Tag D-CYAJI)
Let \(R\) be a ring. An \(R\)-module \(M\) is semisimple if there is a family \((M_j)_{j\in J}\) of simple submodules of \(M\) with \[ M = \bigoplus_{j\in J} M_j. \]
Artin--Wedderburn theorem (Tag T-ZOTWN)
Let \(R\) be a nonzero unital ring that is semisimple as a left module over itself. Then there are integers \(m\geq 1\) and \(n_1,\ldots,n_m\geq 1\) and division rings \(D_1,\ldots,D_m\) such that \[ R \cong \bigoplus_{i=1}^m \Mat_{n_i}(D_i), \] a finite direct sum of the rings of \(n_i\times n_i\) matrices over \(D_i\).
Semisimple rings are finite direct sums of matrix rings over division rings.
Wedderburn's little theorem (Tag T-GYVNQ)
Every finite division ring is commutative, and hence a field.