Find a ring homomorphism $\theta: \mathbb{Z}_6 \times \mathbb{Z}_{14} \to \mathbb{Z}_6 \times \mathbb{Z}_{14}$ for which Ker $\theta = \mathbb{Z}_6 \times \{[0]\}$.
Attempt: I know that Ker $\theta = \{(x, y) \in \mathbb{Z}_6 \times \mathbb{Z}_{14} | \theta (x, y) = (0, 0)\}$. So, I think of $\theta(x, y)=(0, 0)$ (the trivial homomorphism) as one example. Is this example correct?
I am also told that a correct answer is $\theta([x], [y])=([0], [y])$, but I don't understand how this can be true. For instance, $\theta(x, 1)=(0, 1) \notin Ker \theta$. Can someone please explain how this is a correct answer?
Note that we have $(x, 1)\notin \Bbb Z_6\times \{[0]\}$. So in the answer you were given, having $(x, 1)\notin \ker\theta$ seems like it should be a good thing.
Note that if we set $\theta(x, y) = ([0], y)$, in order to get $([0], [0])$, the one and only requirement is that $y = [0]$. So that is the kernel: the subset of $\Bbb Z_6\times \Bbb Z_{14}$ where the second component is $0$. This is exactly $\Bbb Z_6\times \{[0]\}$, which is why the given answer is correct. (It's not the only correct answer, mind you, but it is the simplest one to describe.)