Let $K$ be a field. Show that the map $$f : K \rightarrow K^{2×2}$$ $$ x \mapsto \begin{pmatrix} 1 & x \\ 0 & 1 \\ \end{pmatrix} $$ is injective and that for $x,y \in K: f(x+y)=f(x)\cdot f(y)$.
This task confuses me in its triviality, what do I need to formally prove? Of course $x=y \iff f(x)=f(y)$, because all matrix elements will match. The last part also seems easy, or is there also something to formally be proven there?
The first part of injective property is easily shown by your example of or by noting it is homomorphic and proceeding to show that the kernel is trivial.
The second part is relating to the homomorphism properties of $f$ which is easily shown as a consequence of the dot product, so yes, it is that easy. $$\begin{bmatrix}1 & x \\ 0 & 1\\\end{bmatrix}\cdot\begin{bmatrix}1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & x + y \\ 0 & 1\end{bmatrix}$$ Where matrix multiplication is the operation in $K^{2x2}$ and $K$ is the field with addition.