None:
Polyps:
Strongs:

 Metamath Morphisms / Factor Rings Morphism is a term that covers many variations. The three most common are; Homomorphism Isomorphism Automorphism There are three 'Isomorphism theorems' for rings and their ideals. Before giving them, I'd like to clarify what is meant by doing arithmetic modulo an ideal. An ideal I of a ring R is an additive subgroup of (R , +) and it's order divides that of R by Lagrange's Theorem. We can split R into equivalence classes additively in the form  r + I  (since I is closed, elements of I form the identity  0 + I ). Addition is as simple as one would expect, multiplication is well defined also as, (r + I)(s + I) = rs + rI + sI + rsII = rs + (r + s + rsI)I = rs + I The sub-structure of R defined mod I, is named "The Factor Ring"  R/I. Note A Homomorphism φ on a ring R satisfies for all elements r and s of R, φ( r + s ) = φ( r ) + φ( s ) φ( rs ) = φ( r ) * φ( s ) where * is the multiplication of the image of φ( R ) The first Isomorphism theorem states that any homomorphism φ on a ring R to a Ring S gives rise to a natural isomorphism between;  R / ker(φ) and S,  (that an ideal I is ker(φ))  where ker(φ) is the kernel of the map φ. An isomorphism is a one-to-one homomorphism. If two structures are algebraically identical even if they are described differently, they would be isomorphic is such a map exists. The second Isomorphism relies on the fact that for two ideals I and J of a ring R:   I + J = {i + j : i in I, j in J} IJ = {ij : i in I, j in J} = I ∩ J (the intersection of I and J) Are indeed both ideals. The second isomorphism sates that there is an isomorphism between; (I + J) / J and I / IJ This can be generalised in the case that it will hold also if  I  is only a subring, with J an ideal, though the converse that; (I + J) / I  "is isomorphic to"  J/IJ will not be true. The third isomorphism theorem for Rings states that for two ideals I and J of a ring R, with J a subring and ideal of I, there is an isomorphism between; [R / J] / [I / J] and R / I Because the subfield ideals of a field are trivial, the homomorphisms of a field onto it's subfields require the subfield to be trivial. Since the isomorphism of a field onto a possible rearrangement of itself is the only (interesting) type of homomorphism, these morphisms are termed "Automorphisms" .Continue To Next Page Return To Section Start Return To Previous Page