Artin Wedderburn
1 Current division of team work
Summary: we have defined the necessary structures that are used in the proof. Namely, ideal products, the corner ring, and matrix units. We have proved two major auxiliary theorems that are used in the proof of the main theorem: Brauer’s lemma and theorem 26. Our next target is theorem 25 which is the last prerequisite of the main theorem. After that, we will split the proof of the main theorem into manageable parts.
Matevž Miščič: Has done some major parts of the proof: theorem 6, properties of ideal products and set-like products, properties of corner rings, Brauer’s lemma, theorem 19, theorem 24, and theorem 11. Now he is helping the new members.
Maša Žaucer: Assigned and working on theorem 17 and theorem 25 as an introduction to the project. After that, she will split the proof of the main theorem 28. This will include setting up the definitions of subrings for which the conclusion of the statement holds. A handful of intermediary lemmas will be needed to enable the use of the artinian property.
Job Petrovčič: Has done some major parts of the proof: initial project setup, definition of ideal products, definition of corner ring, basic properties of corner rings, definition of matrix units, and theorem 26. Now he is helping the new members.
If time permits, we will work on the uniqueness part of the proof.
2 Preliminaries
For \(a, b \in R\), denote by \(aRb\) the set \(\{ arb| r \in R\} \).
A left/right ideal \(I\) of a ring \(R\) is an additive subgroup of \(R\) such that \(rI \subseteq I\) for all \(r \in R\) or \(Ir \subseteq I\) for all \(r \in R\), respectively.
A two-sided ideal is a subset of \(R\) that is both left and right ideal of \(R\).
A product of (left/right/two-sided) ideals \(I\) and \(J\) is the ideal \(IJ\) generated by the set of all pairwise products of elements of \(I\) and \(J\).
A ring is prime if we have \(I = 0\) or \(J = 0\) whenever \(IJ = 0\) for some left ideals \(I\) and \(J\).
A ring is prime if and only if for all \(a, b \in R, aRb = 0\) implies \(a = 0\) or \(b = 0\).
\((\Rightarrow )\) Suppose \(aRb = 0\). Then \((Ra)(Rb) = 0\), thus by primality \(Ra = 0\) or \(Rb = 0\). In the former case we get \(a = 1 \cdot a \in Ra\) and thus \(a = 0\) and in the latter case we get \(b = 1 \cdot b \in Rb\) and thus \(b = 0\).
\((\Leftarrow )\) Suppose \(IJ = 0\). Then \(aRb \subseteq IJ = 0\) for any \(a \in I\) and \(b \in J\). By assumption we get \(a = 0\) or \(b = 0\), so at least one of \(I\) and \(J\) is zero.
A ring is prime if and only if for all two-sided ideals \(I\) and \(J\), \(IJ = 0\) implies \(I = 0\) or \(J = 0\).
\((\Rightarrow )\) Two-sided ideals are left ideals, so the result follows directly from definiton.
\((\Leftarrow )\) Suppose \(aRb = 0\). Then \((RaR)(RbR) = 0\). By assumption \(RaR = 0\) or \(RbR = 0\). Thus \(a = 0\) or \(b = 0\) so the result follows from the previous theorem.
A ring is simple if it has no nontrivial two-sided ideals.
A simple ring is prime.
Suppose \(IJ = 0\). If both \(I\) and \(J\) are nonzero, they must be equal to \(R\) by simplicity. But \(RR = R \neq 0\), a contradiction.
Two elements \(a, b \in R\) are orthogonal if \(ab = ba = 0\).
3 Proof of Artin-Wedderburn Theorem for prime and simple rings
The proof is heavily based on [ 1 ] .
If \(e, f \in R\) are orthogonal idempotents and \(f \neq 0\), then the left ideal generated by \(1 - e - f\) is strictly smaller than the left ideal generated by \(1 - e\).
Note that \((1 - e - f)(1 - e) = 1 - e - f\), and hence \(x(1 - e - f) = x(1 - e - f)(1 - e) \in R(1 - e)\) for every \(x \in R\). This proves that \(R(1 - e) \supseteq R(1 - e - f)\).
We have \(f = f(1 - e) \in R(1 - e)\), while \(f = x(1 - e - f)\) with \(x \in R\) implies \(0 = f(1 - f) = x(1 - e - f)(1 - f) = x(1 - e - f) = f\), a contradiction. Therefore, \(R(1 - e) \neq R(1 - e - f)\).
From here on, let \(e\) and \(f\) denote orthogonal idempotents in \(R\).
The set of the corner ring is \(eRe\).
An element \(x\) is in the set \(e R f\) if and only if \(x = e x f\).
\((\Rightarrow )\) Suppose \(x \in e R f\). Then \(x = e y f\) for some \(y \in R\). But then \(e x f = e e y f f = e y f = x\).
\((\Leftarrow )\) Clear.
An element \(x\) of \(R\) is in the corner ring if and only if \(x = e x e\).
Application of theorem 13.
An element \(x\) of the corner ring is of the form \(e y e\) for some \(y \in R\).
Clear from the theorem 14
The corner ring is a (non-unital) subring of \(R\). It has its own unit \(e\).
If \(a, b \in eRe\), then \(a + b = e a e + e b e = e (a + b) e\) so \(eRe\) is closed under addition. If \(a, b \in eRe\), then \(a b = e a e e b e = e a b e\), so \(eRe\) is closed under multiplication. Distributivity and associativity are inherited from \(R\).
Since \(e a = eeae = eae = a = eae = eaee = ae\) for any \(a \in eRe\), \(e\) is the unit of \(eRe\).
If \(R\) is left artinian, then the corner ring is left artinian.
Let \(L_1 \supseteq L_2 \supseteq \ldots \) be a descending chain of left ideals in \(eRe\). Then \( RL_1 \supseteq RL_2 \supseteq \ldots \) is a descending chain of left ideals in \(R\). Since \(R\) is left artinian, this chain stabilizes. But then so does \(eRL_1 \supseteq eRL_2 \supseteq \ldots \). But since \(eRL_i = eReL_i = L_i\), the chain \(L_1 \supseteq L_2 \supseteq \ldots \) also stabilizes.
If \(R\) is a prime ring, then the corner ring is prime.
Suppose \(aeReb = 0\) for \(a, b \in eRe\). Then \(ae = a = 0\) or \(eb = b = 0\) by 6, and by the same theorem, the ring is prime.
If all elements in a ring are left invertible, then the ring is a division ring.
Let \(x \in R\) be arbitrary. Then \(yx = 1\) for some \(y \in R\). Since \(y\) is left invertible, there exists some \(z\) such that \(zy = 1\). By uniqueness of left and right inverses of \(y\) it must hold that \(z = x\). Thus \(x\) is invertible.
Suppose \(L\) is a minimal (left) ideal of \(R\) and \(L^2 \neq 0\). Then there exists an idempotent \(e \in L\) such that \(L = Re\) and \(eRe\) is a division ring.
By assumption, there exists \(y \in L\) such that \(Ly \neq 0\). By minimality \(L = Ly\). Thus, there exists \(e \in L\) such that \(e y = y\). Let \(J \subseteq L\) be the set of elements in \(L\) that annihilate \(y\) from the left.
\(J\) is a left ideal of \(R\) contained in \(L\).
Let \(a, b \in J\). Then \((a + b) y = a y + b y = 0\), so \((a + b) \in J\). For any \(x \in R\), \(x a y = 0\) so \(xa \in J\).
The element \(e\) is not in \(J\), therefore \(J = 0\) by minimality of \(L\). Rearranging the previous equality, \((e^2 - e) y = 0\) which implies \(e^2 = e\), since \(e^2 - e\) is in \(J = 0\). Clearly \(e \neq 0\), and so by minimality \(Re = L\).
Let \(a \in eRe\) be non-zero. Then \(0 \neq Ra = Reae \leq Re = L\), so \(Ra = L\). Thus \(e \in Ra\), so \(e = r a\) for some \(r \in R\). Then \(e = e^2 = e r e a\), so \(a\) is invertible in \(eRe\). We are done by 19
(Already proven in Mathlib) A nonzero left artinian ring has a minimal left ideal.
If minimal left ideal does not exist, then starting with any nonzero left ideal, we can allways find a strictly smaller nonzero left ideal. We can thus construct an infinite strictly decreasing sequence of left ideals, which contradicts the left artinian property.
A set \(e_{ij}\) for \(i, j \in [1, n]\) is a set of matrix units of \(R\) if
and \(\sum _{i=1}^n e_{ii} = 1\).
If \(R\) has a set of matrix units \(e_{ij}\), then \(R\) is isomorphic to the ring of \(n \times n\) matrices over the corner ring \(e_{11}Re_{11}\).
For \(a \in R\), denote \(a_{ij} = e_{1i}ae_{j1}\). Then \(e_{11}a_{ij}e_{11} = e_{11}e_{1i}ae_{j1}e_{11} = e_{1i}ae_{j1}\) by the property of matrix units. Then, the map \(\phi \) claimed to be the isomorphism is \(a \mapsto (a_{ij})_{i,j=1}^n\).
\(\phi \) is additive.
For \(a, b \in R\), we have: \(((a + b)_{ij})_{i,j=1}^n = (e_{1i}(a + b)e_{j1})_{i,j=1}^n = (e_{1i}ae_{j1} + e_{1i}be_{j1})_{i,j=1}^n = (a_{ij} + b_{ij})_{i,j=1}^n\)
The map is multiplicative.
The \((i,j)\) entry of \(\phi (a)\phi (b)\) is equal to
which is the \((i,j)\) entry of \(\phi (ab)\). Therefore, \(\phi (ab) = \phi (a)\phi (b)\).
The map is injective.
Suppose \(a_{ij} = 0\) for all \(i, j\). Then \(e_{ii}ae_{jj} = e_{1i}a_{ij}e_{j1} = 0\). Therefore, \(a = a(\sum _{i=1}^n e_{ii}) = \sum _{i=1}^n ae_{ii} = \sum _{i,j=1}^n e_{ii}ae_{jj} = 0\).
The map is surjective.
Note the \(\phi (e_{k1}ae_{1l})_{kl} = e_{1k}e_{k1}a e_{1l}e_{l1} = e_{11}ae_{11}\) and \(\phi (e_{k1}ae_{1l})_{ab} = e_{1a}e_{k1}a e_{1l}e_{b1} = 0\) if \(a \neq k\) or \(b \neq l\), so \(\phi (e_{k1}ae_{1l})\) is a matrix whose all entries are zero, except the \(k\)-th and \(l\)-th entry is non-zero, and can take arbitrary value in \(e_{11}ae_{11}\). By additivity, the map is surjective.
If a ring \(R\) has a set of pairwise orthogonal idempotents \(e_{ii}\) and
\(e_{1i} \in e_{11}Re_{ii}\) for all \(i\),
\(e_{e1} \in e_{ii}Re_{11}\) for all \(i\),
\(e_{1i}e_{e1} = e_{11}\)
\(e_{i1}e_{1i} = e_{ii}\) for all \(i\),
then \(R\) has matrix units.
Define \(f_{ij} = e_{i1}e_{1j}\).
For \(i = 1\), we have \(f_{1j} = e_{1j}\).
\(f_{1j} = e_{11}e_{1j}\). Since \(e_{1j} \in e_{11}Re_{jj}\), we have \(e_{11}e_{1j} = e_{1j}\) by theorem 13.
For \(j = 1\), we have \(f_{i1} = e_{i1}\) for all \(i\).
\(f_{i1} = e_{i1}e_{11}\). Since \(e_{i1} \in e_{ii}Re_{11}\), we have \(e_{i1}e_{11} = e_{i1}\).
\(f_{1j} f_{k1} = \delta _{jk} f_{11}\) for all \(j, k\)
\(f_{1j} f_{k1} = e_{11}e_{1j}e_{k1}e_{11} = e_{1j}e_{k1} = e_{11} r e_{jj} e_{kk} r' e_{11} = \delta _{jk} e_{11}\) for some \(r, r'\), where the last equality comes from the assumption that the diagonal elements are pairwise orthogonal.
\(f_{ij} f_{kl} = \delta _{jk} f_{il}\).
By definition, \(f_{ij} f_{kl} = e_{i1}e_{1j} e_{k1}e_{1l} = f_{i1}f_{1j} f_{k1}f_{1l} = \delta _{jk} f_{i1} f_{1l} = \delta _{jk} e_{i1} e_{1l} = \delta _{jk} f_{il}\) by the previous claims.
Let \(e, f \in R\) be nonzero orthogonal idempotents and \(R\) a prime ring. Also let \(eRe\) and \(fRf\) be division rings.
Then there exist \(u, v \in R\) such that \(u \in eRf\) and \(v \in fRe\) such that \(uv = e\) and \(vu = f\).
There exists \(a, b \in R\) such that \(eafbe \neq 0\).
Since \(eRe\) is a division ring, there exists \(c \in R\) such that \((eafbe)(ece) = e\). Let \(u = eaf\) and \(v = fbece\), which belong to \(eRf\) and \(fRe\) respectively. Then \(uv = eafbece = e\).
Note that \(vu \in fRf\) and that \(vuv = ve = v = fv\). Therefore, \((vu - f) v = 0\)
\(vu = f\).
Suppose not. Then \(vu - f \neq 0\), but \(vu - f\) is left invertible since \(fRf\) is a division ring. Multiplying by the left inverse, we get \(v = 0 = fv\), a contradiction with the fact that \(uv = e \neq 0\).
If a prime ring \(R\) contains pairwise orthogonal idempotents \(e_{ii}\) with sum \(1\) and \(e_{ii}Re_{ii}\) is a division ring for every \(i\), then \(R\) is isomorphic to \(M_n(e_{11}Re_{11})\).
Applying the theorem 25 for \(e_{11}\) and each \(e_{ii}\), we define \(e_{1i} = u_i\) and \(e_{i1} = v_i\) for each \(i\) wher \(u_i\) and \(v_i\) correspond to \(u\) and \(v\) in the theorem.
By the conclusion of theorem 25, \(e_{1i}e_{i1} = e_{ii}\) and \(e_{i1}e_{1i} = e_{11}\) for all \(i\), and \(e_{1i} \in e_{11}Re_{ii}\) and \(e_{i1} \in e_{ii}Re_{11}\). The \(e_{ii}\) are pairwise orthogonal by assumption.
By theorem 24, \(R\) has matrix units, and by theorem 23 it is isomorphic to \(M_n(e_{11}Re_{11})\).
If \(e, f \in R\) are idempotents and \(f \in (1-e)R(1-e)\) they are orthogonal. Further \(fRf = f(1 - e) R (1 - e) f\).
\(f = f(1 - e) + fe = f + fe\). Thus \(fe = 0\). Similarly, \(ef = 0\). Therefore, \(f\) and \(e\) are orthogonal.
Note that \(x \in fRf \iff \exists r, x = f r f = f \iff \exists r, x = f(1 - e) r (1 - e) f \iff x \in f(1 - e) R (1 - e) f\).
If \(R\) is a prime ring and artinian, then \(R\) is isomorphic to \(M_n(D)\) for some division ring \(D\).
Since \(R\) is artinian, it contains a minimal nonzero left ideal \(L\). If \(L^2 = 0\), this would imply by the prime condition that \(L = 0\), a contradiction. Therefore, \(L^2 \neq 0\). By the Brauer lemma, there exists an idempotent \(e \in L\) such that \(L = Re_{11}\) and \(e_{11}Re_{11}\) is a division ring. By theorem 11 for \(e = 0\) and \(f = e_{11}\) we have that \(R \supsetneq R(1 - e_{11})\).
Suppose \(e_{11} \neq 1\). Then \((1 - e_{11}) R (1 - e_{11})\) is a nonzero ring. It is also prime and artinian by theorems 18 and 17. Repeating the argument for this ring, we obtain \(e_{22}\) such that \(e_{22}(1 - e_{11}) R (1 - e_{11})e_{22}\) is a division ring. Since \(e_{22} \in (1 - e_{11}) R (1 - e_{11})\) then must be orthogonal, as by the theorem 27. Further \(R (1 - e_{11}) \supsetneq R (1 - e_{11} - e_{22})\). Repeating this process, we get a sequence of \(e_{ii}\) and a sequence of left ideals \(R(1 - e_{11} - \ldots - e_{ii})\). By the artinian condition, this sequence must stabilize, so for some \(n\), meaning that \(\sum _{i = 1}^{n} e_{ii} = 1\). \(e_{ii}\) are pairwise orthogonal and are idempotent. Additionally, all \(e_{ii} R e_{ii}\) are division rings. By theorem 26, \(R\) is isomorphic to \(M_n(e_{11}Re_{11})\).
If \(R\) is a simple ring, then \(R\) is isomorphic to \(M_n(D)\) for some division ring \(D\).
Since \(R\) is simple, it is prime. By theorem 28, \(R\) is isomorphic to \(M_n(D)\) for some division ring \(D\).
4 Generalization to semisimple ring
In this section, we prove the following result, which clearly generalizes Artin Wedderburn to semisimple rings
Let \(R\) be a semisimple ring. Then, \(R\) is isomorphic to a direct product of simple, artinian rings.
WLOG, suppose, \(R\) is not simple. We know that \(R\) is (left) artinian, which is a stronger condition that being (two-sided) artinian. Since it is (two-sided) artinian, it must contain a nontrivial minimal (two-sided) ideal \(I\), which is therefore simple. Since \(R\) is semisimple, \(I\) must be a direct summand of \(R\) (AS A LEFT \(R\)-module). Thus, \(R = I \oplus J\) for some (left) ideal \(J\). Then \(1 = i + j\) for some \(i \in I\) and \(j \in J\) . Note that \(I\) and \(J\) are both nontrivial.
\(I J = 0\).
Suppose \(x \in I J\). Then \(x \in I\) since \(I\) is a twosided ideal. Also \(x \in J\) since \(J\) is a left ideal. But then \(x = 0\) since \(I \cap J = 0\).
\(i\) is an idempotent.
\(i = i 1 = i(i + j) = i i + i j = i i\) by the previous claim.
\(I I = I\).
By simplicity of \(I\), \(I I = 0\) or \(I\). Since \( i i = i\), the first case is impossible.
\(J I = 0\).
Note that \(JI\) is spanned by the set of all pairwise products of elements of \(J\) and \(I\). Since \(J\) is a left ideal and \(I\) is a two-sided ideal, \(JI\) is a two-sided ideal. Then it can be either \(0\) or \(I\) by simplicity of \(I\).
Suppose \(JI = I\). Then \(I = I I = I (J I) = (I J) I = 0 \cdot I = 0\), a contradiction.
\(J\) is a two-sided ideal.
We know that it is a left ideal. For arbitrary \(x \in R\), write \(x = x i + x j\). Let \(y \in J\) be arbitrary. Then \(y x = y x i + y x j = 0 + y x j \in J\), where \(y x i = 0\) since it is in \(J I\). Thus \(J\) is also a right ideal.
\( R = I \times J\) as rings.
Let \(x = x_i + x_j\) where \(x_i = x i \in I\) and \(x j \in J\). Similarly, let \(y = y_i + y_j\). Then \(x y = x_i y_i + x_i y_j + x_j y_i + x_j y_j = x_i y_i + x_j y_j\) since \(x_i y_j = 0\) and \(x_j y_i = 0\) by the previous claims. Thus, the map \(x \mapsto (x_i, x_j)\) is a ring homomorphism.
Injective: Suppose \(x i = x_j = 0\). Then \(x = x 1 = x (i + j) = 0\).
Surjective: let \((x_i, x_j)\in I \times J\) be arbitrary. Let \(x = x_i + x_j\). Note that \(x_i = x_i i + x_i j = x_i i\) by orthogonality of \(I\) and \(J\). Similarly \(x_j = x_j j\). Then \(x i = (x_i + x_j) i = x_i i = x_i\) and similarly \(x j = x_j\). Thus \((x_i, x_j)\) is the image of \(x\).
Let \(K \subseteq I\) be a left \(I\)-submodule of \(I\), where \(I\) is treated as a unital ring. Then \(K\) is a left submodule of \(R\).
Let \(r \in R\) and \(k \in K\). Then \(r k = r 1 k = r (i + j) k = r i k + r j k = r i k \in K\) since \(k \in K\), \(r i \in I\) and \(r j k = 0\) as \(j \in J\) and \(k \in I\) and we know \(J I = 0\). Thus \(K\) is closed under left multiplication by element of \(R\).
Both \(I\) and \(J\) are artinian (as rings).
They are both submodules of \(R\) which is assumed to be artinian. Submodules of artinian modules are artinian. Note that \(R\) is artinian since it is semisimple. Thus they are artinian as left \(R\)-modules.
Let \(K_1 \supseteq K_2 \supseteq \ldots \) be a descending chain of left ideals (modules) in ring \(I\). By the previous claim, they are also left ideals in \(R\). Since \(R\) is artinian, this chain stabilizes. But then so does \(K_1 \supseteq K_2 \supseteq \ldots \). Thus \(I\) is artinian. Same argument applies to \(J\).
\(J\) is (left) semisimple.
A submodule of a semisimple module is semisimple. Thus \(J\) is semisimple as a left \(R\)-module. Let \(K \subseteq J\) be a left \(J\)-submodule. Then \(K\) is a left \(R\)-submodule by the previous claim. Since \(R\) is semisimple, every submodules of a submodules has a direct complement, call it \(K'\). Then \(J = K \oplus K'\), as \(R\)-modules. Since \(K'\) is a left \(R\)-submodule, it is also a left \(J\)-submodule. Thus \(J\) is semisimple as a left \(J\)-module.
Thus, we can repeat the process of splitting \(J\) (if it is not simple) into a direct product of simple, artinian rings. Since \(R\) is artinian, this process must stabilize, and we get a direct product of simple, artinian rings.
Apply the 29 to each of the simple, artinian rings to get the desired result.
- 1
Matej Brešar, The Wedderburn-Artin Theorem, arXiv:2405.04588 [math.RA], 2024. https://arxiv.org/abs/2405.04588.