Does there exist a $1-1$ ring homomorphism from $M_d(\mathbb{F})$ to $M_n(\mathbb{F})$ for $d Let $\mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ Then does there exist a injective ring homomorphism from $M_d(\mathbb{F})$ to $M_n(\mathbb{F})$ ? (BTW A ring map sends $1$ to $1$) 
I failed to produce a $1-1$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $19.(b)$ in sec. $13$ (Second Edition). Any help will be appreciated. Thanks.  
Edited Later:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.

Let $K$ be an extension of $F$ of degree $n.$ Prove that $K$ is isomorphic to a subfield of the ring $M_n(F),$ so $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $\leq n.$

 A: As Slade notes in a comment, this is true when $d\mid n$, and the homomorphism from  $M_d(\mathbb F)$ to $M_n(\mathbb F)$ can be defined by sending a $d\times d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.
However, it is not true for general $d$ and $n$. We can see this, for example, with $\mathbb F=\mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
and $B$ be the image of
$$\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3\times 3$ matrix $A$ satisfying $A\ne0$ but $A^2=0$ is, up to similarity,
$$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.
This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $d\not\mid n$, although I expect that it is so.
