Let $X$ be an integral Noetherian scheme, and $\mathcal{K}$ be the constant sheaf with the group $K$ equal to the function field of $X$ where the function field of $X$ is the residue field of generic point.
Then, $\mathcal{K}$ is not coherent unless $X$ is reduced to a point.
I don't understand that meaning of"reduced to a point" and don't know why $\mathcal{K}$ is not coherent.
The theorem says that if $\rm X$ is not the point, then $\mathcal K$ is not a coherent sheaf.
To see this, look at the affine case $\rm X = Spec( A)$ and $\rm K$ is the fraction field of $\rm A$. And try to see when $\rm K$ is a finitely presented $\rm A$-module.
Suppose $\mathcal K$ is coherent (or even just finitely generated over $\mathcal O_X$).
Then for all $x\in X$, the stalk $\mathcal K_x=K$ is finitely generated over $\mathcal O_{X,x}$ and we may apply Nakayama (haha!), which says that $K=0$ if $\mathfrak m_xK=K$ ($\mathfrak m$ being the maximal ideal of $\mathcal O_{X,x}$).
Since this is clearly false we must have $m_xK\neq K$, which implies successively that $\mathfrak m_x=0$, that $\mathcal O_{X,x}$ is a field and that $x$ is the generic point of $X$.
So $X$ indeed has only one point, as claimed.
An $\mathcal O_X$-module $F$ is coherent exactly when $F|_U\cong F(U)^\sim$ for every affine open $U\subset X$, with $F(U)$ finitely generated ($X$ is noetherian) over $\mathcal O_X(U)$. Thus, if $\mathcal K$ were coherent, the function field $K(X)$ of $X$ would have to be a finitely generated $\mathcal O_X(U)$-module (for every $U$), but this is in general not the case because $K(X)=\textrm{Frac }\mathcal O_X(U)$. The only possibility for this to happen is $K(X)=\mathcal O_X(U)$, which means $\mathcal O_{X,\xi}=\bigcap_{x\in U} \mathcal O_{X,x}$ (here $\xi$ is the generic point of $X$). Then, since $\mathcal O_{X,x}\subset \mathcal O_{X,\xi}$ for every $x\in U$, $X$ is one point.
