5.3: Ring Homomorphisms
( \newcommand{\kernel}{\mathrm{null}\,}\)
Ring Homomorphisms
Let R,S be rings. A function ϕ:R→S is called a ring homomorphism if
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ϕ(a+b)=ϕ(a)+ϕ(b),∀a,b∈R, and
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ϕ(ab)=ϕ(a)ϕ(b),∀a,b∈R.
If ϕ is bijective, then ϕ is called isomorphism.
Show that the function ϕ:Z2→Z2 defined by ϕ(x)=x2,x∈Z2 is a ring homomorphism.
Solution
Let a,b∈iZ2.
Consider ϕ(a+b)=(a+b)2=a2+2ab+b2=a2+b2=ϕ(a)+ϕ(b).
Consider ϕ(ab)=(ab)2=a2b2=ϕ(a)ϕ(b).
Hence, ϕ is a ring homomorphism.
Whether or not the function ϕ:Z→Z defined by ϕ(x)=2x,x∈Z is a ring homomorphism? Justify your answer.
Solution
Choose a=2 and b=3. Then ϕ(2)=4 and ϕ(3)=6, ϕ(a)ϕ(b)=(4)(6)=24 but ϕ(ab)=ϕ((2)(3))=ϕ(6)=12≠ϕ(a)ϕ(b).
Hence, ϕ is not a ring homomorphism.
The kernel of ϕ is denoted and defined as Ker(ϕ)={r∈R|ϕ(r)=0}.
Define ϕ:Z→Zn by ϕ(a)=a(modn),∀a∈R.
Then ϕ is a ring homomorphism but not an isomorphism. In this case, Ker(ϕ)=nZ.
Properties of ring homomorphisms:
Let R,S be rings. Let ϕ:R→S be a ring homomorphism. Then,
Then
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ϕ(0)=0.
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if R is a commutative ring, then ϕ(R) is also a commutative ring.
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if R is a field and ϕ(R)≠{0} then ϕ(R) is also a field.
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if R and S have an identity. If ϕ is onto then ϕ(1R)=1S.
Let R,S be rings. Let ϕ:R→S be a ring homomorphism. Then Ker(ϕ) is an ideal of R.