5.3: Ring Homomorphisms
- Page ID
- 132673
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Let \( R, S \) be rings. A function \( \phi: R \to S \) is called a ring homomorphism if
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\( \phi(a+b)=\phi(a)+\phi (b), \forall a,b \in R, \) and
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\( \phi(ab)=\phi(a) \phi (b), \forall a,b \in R. \)
If \( \phi \) is bijective, then \( \phi \) is called isomorphism.
Show that the function \(\phi:\mathbb{Z}_2 \to \mathbb{Z}_2\) defined by \(\phi(x)=x^2, x \in \mathbb{Z}_2 \) is a ring homomorphism.
Solution
Let \(a, b \in i\mathbb{Z}_2.\)
Consider \( \phi(a+b)=(a+b)^2=a^2+2ab+b^2=a^2+b^2=\phi(a)+\phi (b).\)
Consider \( \phi(ab)=(ab)^2=a^2b^2=\phi(a)\phi (b).\)
Hence, \(\phi\) is a ring homomorphism.
Whether or not the function \(\phi:\mathbb{Z} \to \mathbb{Z}\) defined by \(\phi(x)=2x, x \in \mathbb{Z} \) is a ring homomorphism? Justify your answer.
Solution
Choose \(a=2\) and \(b=3\). Then \( \phi(2)= 4\) and \(\phi(3)=6\), \(\phi(a)\phi (b)=(4)(6)=24\) but \( \phi(ab)=\phi((2)(3))=\phi(6)=12\ne \phi(a)\phi(b)\).
Hence, \(\phi\) is not a ring homomorphism.
The kernel of \( \phi \) is denoted and defined as \( Ker(\phi) = \{ r \in R | \phi(r)=0 \}. \)
Define \( \phi: \mathbb{Z} \to \mathbb{Z}_n \) by \( \phi(a)=a(mod n), \forall a\in R. \)
Then \( \phi \) is a ring homomorphism but not an isomorphism. In this case, \( Ker(\phi) = n \mathbb{Z}. \)
Properties of ring homomorphisms:
Let \( R, S \) be rings. Let \( \phi: R \to S \) be a ring homomorphism. Then,
Then
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\( \phi(0)=0. \)
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if \( R \) is a commutative ring, then \( \phi(R) \) is also a commutative ring.
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if \( R \) is a field and \( \phi(R) \ne \{0\} \) then \( \phi(R) \) is also a field.
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if \( R \) and \(S \) have an identity. If \( \phi \) is onto then \( \phi(1_R)=1_S. \)
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Let \( R, S \) be rings. Let \( \phi: R \to S \) be a ring homomorphism. Then \( Ker (\phi)\) is an ideal of \(R\).