Intuition behind $k$-algebra, $k$-algebra morphisms? I will state the definition of a $k$-algebra and $k$-algebra morphisms.

A ring $A$ equipped with a ring homomorphism $k \to Z(A)$ is called a $k$-algebra. More explicitly, this means that $A$ has a structure of vector space over $k$ and also a ring structure such that:
(1) The operations "$+$" coming from the vector space structure and the ring structure, respectively, are the same.
(2) The ring multiplication $\cdot$: $A \times A \to A$ is a $k$-bilinear map.
One defines $k$-algebra morphisms as $k$-linear ring morphisms.

I understand these definitions formally, but I have quite a poor grasp on the intuition behind the definition of $k$-algebra and $k$-algebra morphisms. Could someone provide me with their intuitions on such?
 A: If $k$ is a field and $A$ is a unital $k$-algebra with unit $\mathbb{1}$, then $A$ is not only a $k$-vector space, but the scalars can be thought of as elements of $A$: the scalar $\lambda\in k$ can also be thought of as the scalar multiple $\lambda \mathbb{1}$ of the unit. A $k$-algebra homomorphism $\varphi : A \to B$ between $k$-algebras $A$ and $B$ is a ring homomorphism such that $\varphi(\lambda\mathbb{1}_A) = \lambda\mathbb{1}_B$ holds, that is, scalars in $A$ are sent to their corresponding scalars in $B$. This serves to rule out "strange" ring homomorphisms that don't preserve the $k$-algebra structure. For instance, consider the following example:

Example. The polynomial ring $\mathbb{C}[x]$ is a $\mathbb{C}$-algebra in a natural way. Now consider the ring homomorphism $\psi : \mathbb{C}[x] \to \mathbb{C}[y]$ given by $\lambda\cdot x^n \mapsto \overline{\lambda}\cdot y^n$. This is not a $\mathbb{C}$-algebra homomorphism: for instance we have $\psi(i) = -i$.

In general, if $A$ is a unital algebra over a commutative unital ring $R$ and $\alpha\in A$ is an element, then there is a unique unital $R$-algebra homomorphism $\varphi : R[x] \to A$ such that $\varphi(x) = \alpha$ holds. (If $p\in R[x]$ is a polynomial, then $\varphi(p)$ is simply the polynomial $p$ applied to $\alpha$.) The example above shows that there may be different ring homomorphisms with that same property (in the case $A = \mathbb{C}[y]$ and $\alpha = y$) that are not $k$-algebra homomorphisms.
In the study of non-unital algebras, a different definition of a $k$-algebra is customary. Now a $k$-algebra over a field $k$ is defined to be a $k$-vector space $A$ together with a multiplication such that $(A,+,\cdot)$ forms a rng (that is, a ring without unit). In this case the scalars are not in general elements of $A$, and the algebra structure cannot be defined in terms of a homomorphism $k \to Z(A)$. Once again, a $k$-algebra homomorphism $\varphi : A \to B$ between $k$-algebras $A$ and $B$ is a linear map that preserves multiplication: $\varphi(xy) = \varphi(x)\varphi(y)$ for all $x,y\in A$, though we cannot require $\varphi$ to be unital. Regardless of whether the algebra is unital, I find it easier to think of a $k$-algebra as a vector space equipped with multiplication instead of a ring together with a homomorphism. (However if $k$ is not a field but an arbitrary ring, then a $k$-algebra is no longer a vector space.)
