Find kernel of a homomorphism $f$

Question

I have $f: \mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ , $f$ is homomorphism such that $f(1,0) = (2,-1)$ and $f(0,1) = (-6,5)$. Find kernel of $f$. My progress:

$\ker(f)=\{(a,b)\in \mathbb{Z}^2: f(a,b)=(0,0) \}$

$f(a,b)=f(a,0)+f(0,b)=af(1,0)+bf(0,1)=a(2,-1)+b(-6,5)=(2a-6b,-a+5b) = (0,0)$

Now I get $2a-6b=-a+5b \Rightarrow b = 3/11a$. However, this seems to be wrong:

$b = 3/11a \Rightarrow (-4/11a,4/11a)=(0,0)$. If $a=11$, $(−4/11a,4/11a)\neq (0,0)$.

So, what is wrong here?

Answer 1

$$f(a,b) = (0,0)\iff (2a-6b, 5b-a) = (0,0)$$ $$\iff a = 3b, a = 5b$$ $$\iff a = b = 0$$

Hence, $\ker f = \{(0,0)\}$ and $f$ is injective.

Answer 2

The homomorphism $f$ is injective because $$\det f=\begin{vmatrix}2&-6\\-1&5\end{vmatrix}=4\ne 0.$$

More generally, over a commutative ring $R$, an endomorphism of a free $R$-module is injective if and only if $\det f$ is a non-zero divisor – which means nonzero if $R$ is an integral domain.