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

Printable View

- May 6th 2009, 03:05 AMMaccamanRing homomorphism
Let $\displaystyle f : \mathbb{Z}/3\mathbb{Z} \rightarrow \ \mathbb{Z}/6\mathbb{Z} $ be defined by $\displaystyle f(x) = 4x $ Is $\displaystyle f $ a well-defined ring homomorphism? If so, find its kernel and image.

- May 6th 2009, 03:14 AMNonCommAlg
yes it is and the kernel is (0). the easiest way to see this is to just check 3 elements of $\displaystyle \mathbb{Z}/3\mathbb{Z}$ and see that $\displaystyle f(x)=0$ iff $\displaystyle x=0.$ also you need to check that $\displaystyle f(xy)=f(x)f(y),$ which is because

$\displaystyle 16=4$ in $\displaystyle \mathbb{Z}/6\mathbb{Z}.$ the image is $\displaystyle \{0,2,4 \}.$ (here by 0 i mean the coset $\displaystyle 0 + 6 \mathbb{Z},$ and by 2 i mean $\displaystyle 2 + 6\mathbb{Z},$ etc.) - May 12th 2009, 04:50 AMMaccaman
Thank-you for your help, but I'm still confused because I'm getting things mixed up. I'm having trouble seeing that $\displaystyle f $ is a well-defined ring homomorphism to start with. Could you show me why it is?? (Worried)