It would have been better if you had given some context. Reduced norm of what? A quaternion algebra? A division algebra? A central simple algebra?
Let's say you're talking about a central simple algebra $A$ over a field $K$ of dimension $n^2$. Then there's always some extension of $L$ of $K$ such that $A\otimes_K L$ is isomorphic to the matrix algebra $M_n(L)$. In particular tensoring the field injection $K \hookrightarrow L$ with $A$ over $K$ gives an embedding $A \hookrightarrow M_n(L)$. This is convenient because it allows us to represent the algebra $A$ as matrices. The reduced norm (and reduced trace) are then the determinant (and trace) of the matrix associated to an element of $A$. In particular, they don't depend on the choice of the splitting field $L$.
Taking the determinant of a $n\times n$ matrix $X$ is much more natural than taking the determinant of the $n^2 \times n^2$ matrix which multiplication by $X$ induces on the vector space of $n\times n$ matrices.
Consider a quaternion algebra consisting of elements $x=\alpha+i\beta+j\gamma+k\delta$. Then there's a natural involution $x\mapsto x^t = \alpha - i\beta -j \gamma -k \delta$ and the reduced norm of $x$ is $xx^t$, the reduced trace is $x+x^t (=2\alpha)$. As you can check they're easy to compute and very natural. if $i^2=j^2=k^2=-1$ we have $xx^t = \alpha^2 + \beta^2 + \gamma^2 + \delta^2$, so for Hamilton's quaternions the reduced norm is also a generalization of the (number-theoretic) norm of a complex number: $N(x+iy) = (x+iy)(x-iy) = x^2 + y^2$.
For field extensions we use the regular norm and trace more often, but for say, quaternion algebras, we almost always use reduced norm and reduced trace instead. Aside from the invariance property and ease of computation, finding splitting fields for a central simple algebra is an important aspect of the classification theory, so something we end up doing quite often anyway.