Ring homomorphism

**Maccaman** · May 6th 2009, 03:05 AM

Let $f : \mathbb{Z}/3\mathbb{Z} \rightarrow \ \mathbb{Z}/6\mathbb{Z}$ be defined by $f(x) = 4x$ Is $f$ a well-defined ring homomorphism? If so, find its kernel and image.

**NonCommAlg** · May 6th 2009, 03:14 AM

Originally Posted by Maccaman

Let $f : \mathbb{Z}/3\mathbb{Z} \rightarrow \ \mathbb{Z}/6\mathbb{Z}$ be defined by $f(x) = 4x$ Is $f$ a well-defined ring homomorphism? If so, find its kernel and image.

yes it is and the kernel is (0). the easiest way to see this is to just check 3 elements of $\mathbb{Z}/3\mathbb{Z}$ and see that $f(x)=0$ iff $x=0.$ also you need to check that $f(xy)=f(x)f(y),$ which is because

$16=4$ in $\mathbb{Z}/6\mathbb{Z}.$ the image is $\{0,2,4 \}.$ (here by 0 i mean the coset $0 + 6 \mathbb{Z},$ and by 2 i mean $2 + 6\mathbb{Z},$ etc.)

**Maccaman** · May 12th 2009, 04:50 AM

Thank-you for your help, but I'm still confused because I'm getting things mixed up. I'm having trouble seeing that $f$ is a well-defined ring homomorphism to start with. Could you show me why it is??

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