Let $\det\_d = \det((x\_{i,j})\_{1 \leq i,j\leq d})$ be the determinant of a generic $d \times d$ matrix. Suppose $k \mid d$, $1 < k < d$. Can $\det\_d$ be written as the determinant of a $k \times k$ matrix of forms of degree $d/k$?


Even writing $\det\_4$ as the determinant of a $2 \times 2$ matrix of quadratic forms seems impossible, just intuitively. The space of $2 \times 2$ matrices of quadratic forms in $16$ variables has dimension $4 \cdot \binom{17}{2} = 544$, while the space of quartics in $16$ variables has dimension $\binom{19}{4} = 3876$.



Is the determinant equal to a determinant?
I think a result of Hochster allows to get a quick proof that it is not possible to express the determinant of the generic $d \times d$ matrix as the determinant of a $k \times k$ matrices with entries being homogeneous forms of degree $\dfrac{d}{k}$, provided that $1 <k <d$.


I will work over an algebraically closed field.
Relying on some cohomological methods (similar to the ones Jason Starr is using in his answer), Hochster proved that the variety of $n \times n$ matrices with $\textrm{rank} < n-1$ can be set-theoretically defined by $2n$ equations **and not less** (see this paper by Bruns and Schwänzl for some improvement of Hochster's result : <http://www.home.uni-osnabrueck.de/wbruns/brunsw/pdf-article/NumbEqDet.published.pdf>).


Now, we proceed by absurd. Assume that the determinant of the generic $d \times d$ matrix can be written as the determinant of $k \times k$ matrix (say $A$) with forms homogeneous of degree $\dfrac{d}{k}$. We assume $1 < k < d$. We denote by $P\_1, \ldots, P\_{k^2}$ the entries $A$, which are homogeneous polynomials in $x\_1, \ldots, x\_{d^2}$ of degree $\dfrac{d}{k}$.


Let $B$ be the generic $k \times k$ matrix with entries $Y\_1, \ldots, Y\_{k^2}$. Denote by $Q\_1, \ldots Q\_{k^2}$ the $k-1$ minors of $B$. The variety defined by the vanishing of the $\{Q\_i\}\_{i=1\ldots k^2}$ is **non-empty** of codimension $4$ in $\mathbb{A}^{k^2}$ (here I use that $k>1$). Hence, if we replace $Y\_i$ by $P\_i(x\_1, \ldots, x\_{d^2})$, we see that the scheme defined by the vanishing of the $\{Q\_i(P\_1,\ldots,P\_{k^2})\}\_{i=1 \ldots k^2}$ has codimension at most $4$ in $\mathbb{A}^{d^2}$.


Furthermore, a simple computation of partial derivatives shows that the scheme defined by the vanishing of the $\{Q\_i(P\_1,\ldots,P\_{k^2})\}\_{i=1 \ldots k^2}$ is included in the singular locus of the variety defined by $\det A = 0$. But $\det A$ is the determinant of the generic $d \times d$ matrix, so that its singular locus is the variety of matrices of $\textrm{rank} < d-1$ : it is irreducible of codimension $4$ in $\mathbb{A}^{d^2}$.


From the above, we deduce that the variety of $d \times d$ matrices of $\textrm{rank} < d-1$ is set-theoretically equal to the scheme defined by the
 $\{Q\_i(P\_1,\ldots,P\_{k^2})\}\_{i=1 \ldots k^2}$. 


By Hochster's result (the existence part), one can find $2k$ polynomials (say $T\_1, \ldots, T\_{2k}$) in the ideal generated by $Q\_1, \ldots, Q\_{k^2}$ such that $$\textrm{rad}(T\_1, \ldots, T\_{2k}) = \textrm{rad}(Q\_1, \ldots, Q\_{k^2}).$$ 


Replacing $Y\_i$ by $P\_i(x\_1,\ldots, x\_{d^2})$ in the $\{T\_j\}\_{j=1 \ldots 2k}$, we find that the vanishing of the $\{T\_j(P\_1, \ldots, P\_{k^2}) \}\_{j=1 \ldots 2k}$ defines set-theoretically the variety of $d \times d$ matrices of $\textrm{rank} < d-1$. Since $2k < 2d$, we get a contradiction with Hochster's result.


