In ring theory, a branch of mathematics, the Skolem–Noether theorem characterizes the automorphisms of simple rings. It is a fundamental result in the theory of central simple algebras.
 The theorem was first published by Thoralf Skolem in 1927 in his paper Zur Theorie der assoziativen Zahlensysteme (German: On the theory of associative number systems) and later rediscovered by Emmy Noether.
 In a general formulation, let A and B be simple unitary rings, and let k be the center of B. The center k is a field since given x nonzero in k, the simplicity of B implies that the nonzero two-sided ideal BxB = (x) is the whole of B, and hence that x is a unit. If the dimension of B over k is finite, i.e. if B is a central simple algebra of finite dimension, and A is also a k-algebra, then given k-algebra homomorphisms
 there exists a unit b in B such that for all a in A
 In particular, every automorphism of a central simple k-algebra is an inner automorphism.
 First suppose 
  
    
      
        B
        =
        
          M
          
            n
          
        
        ⁡
        (
        k
        )
        =
        
          End
          
            k
          
        
        ⁡
        (
        
          k
          
            n
          
        
        )
      
    
    {\displaystyle B=\operatorname {M} _{n}(k)=\operatorname {End} _{k}(k^{n})}
  
. Then f and g define the actions of A on 
  
    
      
        
          k
          
            n
          
        
      
    
    {\displaystyle k^{n}}
  
; let 
  
    
      
        
          V
          
            f
          
        
        ,
        
          V
          
            g
          
        
      
    
    {\displaystyle V_{f},V_{g}}
  
 denote the A-modules thus obtained. Since 
  
    
      
        f
        (
        1
        )
        =
        1
        ≠
        0
      
    
    {\displaystyle f(1)=1\neq 0}
  
 the map f is injective by simplicity of A, so A is also finite-dimensional. Hence two simple A-modules are isomorphic and 
  
    
      
        
          V
          
            f
          
        
        ,
        
          V
          
            g
          
        
      
    
    {\displaystyle V_{f},V_{g}}
  
 are finite direct sums of simple A-modules. Since they have the same dimension, it follows that there is an isomorphism of A-modules 
  
    
      
        b
        :
        
          V
          
            g
          
        
        →
        
          V
          
            f
          
        
      
    
    {\displaystyle b:V_{g}\to V_{f}}
  
. But such b must be an element of 
  
    
      
        
          M
          
            n
          
        
        ⁡
        (
        k
        )
        =
        B
      
    
    {\displaystyle \operatorname {M} _{n}(k)=B}
  
. For the general case, 
  
    
      
        B
        
          ⊗
          
            k
          
        
        
          B
          
            op
          
        
      
    
    {\displaystyle B\otimes _{k}B^{\text{op}}}
  
 is a matrix algebra and that 
  
    
      
        A
        
          ⊗
          
            k
          
        
        
          B
          
            op
          
        
      
    
    {\displaystyle A\otimes _{k}B^{\text{op}}}
  
 is simple. By the first part applied to the maps 
  
    
      
        f
        ⊗
        1
        ,
        g
        ⊗
        1
        :
        A
        
          ⊗
          
            k
          
        
        
          B
          
            op
          
        
        →
        B
        
          ⊗
          
            k
          
        
        
          B
          
            op
          
        
      
    
    {\displaystyle f\otimes 1,g\otimes 1:A\otimes _{k}B^{\text{op}}\to B\otimes _{k}B^{\text{op}}}
  
, there exists 
  
    
      
        b
        ∈
        B
        
          ⊗
          
            k
          
        
        
          B
          
            op
          
        
      
    
    {\displaystyle b\in B\otimes _{k}B^{\text{op}}}
  
 such that
 for all 
  
    
      
        a
        ∈
        A
      
    
    {\displaystyle a\in A}
  
 and 
  
    
      
        z
        ∈
        
          B
          
            op
          
        
      
    
    {\displaystyle z\in B^{\text{op}}}
  
. Taking 
  
    
      
        a
        =
        1
      
    
    {\displaystyle a=1}
  
, we find
 for all z. That is to say, b is in 
  
    
      
        
          Z
          
            B
            ⊗
            
              B
              
                op
              
            
          
        
        (
        k
        ⊗
        
          B
          
            op
          
        
        )
        =
        B
        ⊗
        k
      
    
    {\displaystyle Z_{B\otimes B^{\text{op}}}(k\otimes B^{\text{op}})=B\otimes k}
  
 and so we can write 
  
    
      
        b
        =
        
          b
          ′
        
        ⊗
        1
      
    
    {\displaystyle b=b'\otimes 1}
  
. Taking 
  
    
      
        z
        =
        1
      
    
    {\displaystyle z=1}
  
 this time we find
 which is what was sought.