Thread: Algebra isomorphism View Single Post
 Sci Advisor HW Helper P: 9,453 Oh I guess morphism's original point of view makes it clear how to write down the inverse. I.e. he shows the isomorphism is essentially given by the "chinese remainder theorem". So if one can write down an inverse to the map Z/(mn)-->Z/m x Z/n, where gcd(m,n) = 1, then one should be able to write down an inverse to: C[x]/(x^2+1)=C[x]/(x+i)⊕C[x]/(x−i).