Mohammad,

flatness is not simple, so you are not going to get a simple overall definition. On the other hand, in the example you mention in your comment, there is a simple criterion:

Let $f:X\to Y$ be a morphism of schemes such that $X$ is reduced and irreducible (most likely satisfied in the cases you are interested in, at least for now) and $Y$ is a smooth curve. Then if $X$ dominates $Y$ (i.e., the image is dense in $Y$) then $f$ is flat.

In particular $f:\mathbb C^2\to \mathbb C$ given by $(x,y)\mapsto xy$ is flat.

Over higher dimensional basis it is a little more complicated, but there is one simple criterion you can check:

If $f$ is flat, then the fibers have the same dimension.

For example a blowup is the typical not flat map (think of the meaning of "flat").

On the other hand, the fibers being equidimensional is certainly not enough for a morphism to be flat. An example is provided by taking $X$ to be the union of two planes meeting in a single point mapping to just one copy of the plane in the obvious way. All fibers are $0$-dimensional, but the morphism is not flat.

However, there is a criterion that is pretty useful and says that if both $X$ and $Y$ are relatively nice, then flatness is equivalent to that.

Let $f:X\to Y$ be a morphism of schemes such that $X$ is Cohen-Macaulay and $Y$ is regular (smooth if you prefer), then $f$ is flat if and only if the fibers are equidimensional.

Finally, my suggestion is that in order to get a better feeling for flatness, try looking at finite morphisms as in the above example. If you understand those, you essentially have understood all flat maps.

Mohammad,

flatness is not simple, so you are not going to get a simple overall definition. On the other hand, in the example you mention in your comment, there is a simple criterion:

Let $f:X\to Y$ be a morphism of schemes such that $X$ is reduced and irreducible (most likely satisfied in the cases you are interested in, at least for now) and $Y$ is a smooth curve. Then if $X$ dominates $Y$ (i.e., the image is dense in $Y$) then $f$ is flat.

In particular $f:\mathbb C^2\to \mathbb C$ given by $(x,y)\mapsto xy$ is flat.

Over higher dimensional basis it is a little more complicated, but there is one simple criterion you can check:

If $f$ is flat, then the fibers have the same dimension.

For example a blowup is the typical not flat map (think of the meaning of "flat").

On the other hand, the fibers being equidimensional is certainly not enough for a morphism to be flat. An example is provided by taking $X$ to be the union of two planes meeting in a single point mapping to just one copy of the plane in the obvious way. All fibers are $0$-dimensional, but the morphism is not flat.

However, there is a criterion that is pretty useful and says that if both $X$ and $Y$ are relatively nice, then flatness is equivalent to that.

Let $f:X\to Y$ be a dominant morphism of irreducible varieties such that $X$ is Cohen-Macaulay and $Y$ is regular (smooth if you prefer), then $f$ is flat if and only if the fibers are equidimensional.

Finally, my suggestion is that in order to get a better feeling for flatness, try looking at finite morphisms as in the above example. If you understand those, you essentially have understood all flat maps.