Integral domain which contains a copy of its fraction field

Question

Let $R$ be an integral domain with fraction field $K$. If there exists an injective ring homomorphism from $K$ to $R$ , then is it true that $R$ is a field ?

Strictly speaking, I am not saying $K \subseteq R$ and the injective map say $\phi : K \to R$ could possibly be anything, so I am not quite sure whether this really does imply $R$ is a field or not (I feel it should ... )

Answer 1

Let $(x_n)_{n\ge0}$ be a sequence of algebraically independent indeterminates over the field $F$ and let $K=F(x_1,x_2,\dots,x_n,\dotsc)$. Consider $R=K[x_0]$, which is not a field. Its fraction field is $F(x_0,x_1,\dots,x_n,\dotsc)\cong K$.