Suppose you want to do moduli theory or to put it simpler, you are interested in deformations and degenerations. Often the degenerate objects have a natural non-reduced structure. In fact it is possible that taking the corresponding reduced scheme screws things up.

Here are two simple examples:

Example #1: Consider the morphism $\mathbb A^2\to \mathbb A^1$ defined by $(x,y)\mapsto x^2$. The fibers are the curves defined by $x^2=\lambda$. For $\lambda\neq 0$ this is a parabola and for $\lambda=0$ a (double) line. If we only consider reduced schemes, then this is just a line, but otherwise we would expect that the members of a family of plane curves have the same intersection numbers (counted properly and also counting intersections at infinity) with other curves. Taking another line in general position one can see easily that the parabola intersects it in $2$ points while the line in only $1$. Considering the scheme theoretic fiber $x^2=0$ which is a double line resolves this problem.

Example #2: Let $X=\{(1,\lambda t, t^2,t^3)\vert (t,\lambda)\in \mathbb A^2\}\subset \mathbb A^3$. This is a surface defined "classically". Consider its projection to $\mathbb A^1$ by mapping the point $(1,\lambda t, t^2,t^3)$ to $\lambda$. Denote this by $f:X\to\mathbb A^1$. Still pretty classical. Now notice that the (classical=reduced) fiber of $f$ over $\lambda=0$ is a nodal cubic curve while for $\lambda\neq 0$ its it is a twisted cubic. Also notice that this family can easily be compactified to be a projective family, so we get a family of $\mathbb P^1$'s degenerating to a projective nodal curve. However, without nilpotents this leads to severe headache.

Since $X$ is irreducible and $\mathbb A^1$ is non-singular, $f$ should be flat. But fibers of a flat morphism have constant Hilbert polynomials, in particular their arithmetic genus is constant. The arithmetic genus of a twisted cubic (i.e., $\mathbb P^1$) is $0$ while that of a nodal cubic is $1$. If you want a completely classical argument, then one could say that the nodal cubic is also an obvious degeneration of non-singular plane cubic curves. Working over the complex numbers a plane cubic is a torus while a twisted cubic is a sphere. So this would suggest that it is possible to deform a sphere to a torus....

The resolution of this dilemma is that if you compute the scheme theoretic fiber (using fibre products) then you'll see that the correct fibre over $\lambda=0$ is actually the nodal cubic, with a nilpotent sitting at the singularity. So the fibre is a non-reduced scheme and its arithmetic genus is $0$ so we can all sleep peacefully.

2 added 621 characters in body

Let me add an answer that is similar to the other answers but may have a slightly different point of view.

Suppose you want to do moduli theory or to put it simpler, you are interested in deformations and degenerations. Often the degenerate objects have a natural non-reduced structure. In fact it is possible that taking the corresponding reduced scheme screws things up.

Here are two simple examples:

Example #1: Consider the morphism $\mathbb A^2\to \mathbb A^1$ defined by $(x,y)\mapsto x^2$. The fibers are the curves defined by $x^2=\lambda$. For $\lambda\neq 0$ this is a simple exampleparabola and for $\lambda=0$ a (double) line. If we only consider reduced schemes, then this is just a line, but otherwise we would expect that the members of a family of plane curves have the same intersection numbers (counted properly and also counting intersections at infinity) with other curves. Taking another line in general position one can see easily that the parabola intersects it in $2$ points while the line in only $1$. Considering the scheme theoretic fiber $x^2=0$ which is a double line resolves this problem.

Example #2: Let $X=\{(1,\lambda t, t^2,t^3)\vert (t,\lambda)\in \mathbb A^2\}\subset \mathbb A^3$. This is a surface defined "classically". Consider its projection to $\mathbb A^1$ by mapping the point $(1,\lambda t, t^2,t^3)$ to $\lambda$. Denote this by $f:X\to\mathbb A^1$. Still pretty classical. Now notice that the (classical=reduced) fiber of $f$ over $\lambda=0$ is a nodal cubic curve while for $\lambda\neq 0$ its a twisted cubic. Also notice that this family can easily be compactified to be a projective family, so we get a family of $\mathbb P^1$'s degenerating to a projective nodal curve. However, without nilpotents this leads to severe headache.

Since $X$ is irreducible and $\mathbb A^1$ is non-singular, $f$ should be flat. But fibers of a flat morphism have constant Hilbert polynomials, in particular their arithmetic genus is constant. The arithmetic genus of a twisted cubic (i.e., $\mathbb P^1$) is $0$ while that of a nodal cubic is $1$.

The resolution of this dilemma is that if you compute the scheme theoretic fiber (using fibre products) then you'll see that the correct fibre over $\lambda=0$ is actually the nodal cubic, with a nilpotent sitting at the singularity. So the fibre is a non-reduced scheme and its arithmetic genus is $0$ so we can all sleep peacefully.

1

Let me add an answer that is similar to the other answers but may have a slightly different point of view.

Suppose you want to do moduli theory or to put it simpler, you are interested in deformations and degenerations. Often the degenerate objects have a natural non-reduced structure. In fact it is possible that taking the corresponding reduced scheme screws things up.

Here is a simple example:

Let $X=\{(1,\lambda t, t^2,t^3)\vert (t,\lambda)\in \mathbb A^2\}\subset \mathbb A^3$. This is a surface defined "classically". Consider its projection to $\mathbb A^1$ by mapping the point $(1,\lambda t, t^2,t^3)$ to $\lambda$. Denote this by $f:X\to\mathbb A^1$. Still pretty classical. Now notice that the (classical=reduced) fiber of $f$ over $\lambda=0$ is a nodal cubic curve while for $\lambda\neq 0$ its a twisted cubic. Also notice that this family can easily be compactified to be a projective family, so we get a family of $\mathbb P^1$'s degenerating to a projective nodal curve. However, without nilpotents this leads to severe headache.

Since $X$ is irreducible and $\mathbb A^1$ is non-singular, $f$ should be flat. But fibers of a flat morphism have constant Hilbert polynomials, in particular their arithmetic genus is constant. The arithmetic genus of a twisted cubic (i.e., $\mathbb P^1$) is $0$ while that of a nodal cubic is $1$.

The resolution of this dilemma is that if you compute the scheme theoretic fiber (using fibre products) then you'll see that the correct fibre over $\lambda=0$ is actually the nodal cubic, with a nilpotent sitting at the singularity. So the fibre is a non-reduced scheme and its arithmetic genus is $0$ so we can all sleep peacefully.