Question

If \(\phi:\mathbb{Z}_7\to \mathbb{Z}_{12}\) is defined such that \(\phi(a)=4a\), then \(\phi\) is

• A. a ring homomorphism
• B. not a ring homomorphism
• C. a ring homomorphism but not isomorphism
• D. Isomorphism

Hint:

For maps between modular rings, always check whether the definition is compatible with the modulus relation of the domain.

Answer 1

The Correct Option is B

Concept:
A ring homomorphism must preserve addition and multiplication. That means for all \(a,b\), \[ \phi(a+b)=\phi(a)+\phi(b) \] and \[ \phi(ab)=\phi(a)\phi(b) \]

Step 1: Write the mapping. \[ \phi:\mathbb{Z}_7\to\mathbb{Z}_{12} \] is defined by \[ \phi(a)=4a \] The value is taken modulo \(12\) in the codomain.

Step 2: Test multiplicative property. Take \[ a=1,\qquad b=1 \] Then, \[ ab=1 \] So, \[ \phi(ab)=\phi(1)=4 \] Now, \[ \phi(1)\phi(1)=4\cdot 4 \] \[ =16 \] In \(\mathbb{Z}_{12}\), \[ 16\equiv 4\pmod{12} \] So this example does not fail. Now take \[ a=2,\qquad b=2 \] Then in \(\mathbb{Z}_7\), \[ ab=4 \] So, \[ \phi(ab)=\phi(4)=16\equiv 4\pmod{12} \] But \[ \phi(2)\phi(2)=(8)(8)=64 \] In \(\mathbb{Z}_{12}\), \[ 64\equiv 4\pmod{12} \] This also matches. However, for a ring homomorphism from a field \(\mathbb{Z}_7\) to \(\mathbb{Z}_{12}\), the additive structure must be compatible with the moduli.

Step 3: Check additive compatibility with zero. In \(\mathbb{Z}_7\), \[ 7\cdot 1=0 \] For a ring homomorphism, this should imply \[ 7\phi(1)=0 \] But \[ \phi(1)=4 \] So, \[ 7\phi(1)=7\cdot 4=28 \] In \(\mathbb{Z}_{12}\), \[ 28\equiv 4\pmod{12} \] which is not \(0\). Therefore, addition is not properly preserved. Hence \(\phi\) is not a ring homomorphism.

Step 4: Final answer. \[ \boxed{\text{not a ring homomorphism}} \]