Is the determinant equal to a determinant? 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 $\binom{19}{4} = 3876$.
 A: The answer is that if $1 < k < d$, then such a matrix is equivalent to a diagonal matrix with $\det(x_{ij})=: D$ on the diagonal and the rest of the diagonal entries equal to $1$. This does not require that $k$ divides $d$. 
The reason is the following: If $M$ is a $k\times k$ matrix with entries from the polynomial ring $S$ in the $x_{ij}$ over some field and $\det M = D$, then the cokernel of $M$, viewed as an $S$-linear map from $S^k$ to $S^k$  defines a maximal Cohen--Macaulay module of rank one over the hypersurface ring $R=S/(D)$, generated by at most $k$ elements. For this statement see Eisenbud's paper MR0570778 (82d:13013). 
In the book by Bruns and Vetter on determinantal rings MR0953963 (89i:13001) all reflexive modules of rank one over a determinantal ring are determined and in the situation here they show that only the cokernel of the generic matrix itself or its transpose gives a nontrivial maximal Cohen-Macaulay module of rank one. 
However these need at least $d$ generators.
The trivial module is just $R$ itself and in terms of matrices it means that $k<d$ implies that $M$ is equivalent to a diagonal matrix with $D$ on the diagonal and the rest of the diagonal filled up by $1$'s. (The case I neglected in my original posting...) 
Another interesting source for similar questions (and some answers) is the article by G. Bergman MR2205070 (2006k:15037).
A: I will work over a field of characteristic $0$ so that reductive algebraic groups are linearly reductive; presumably there is a way to eliminate this hypothesis.  In this case, for an integer $n>1$ and for a divisor $m$ such that $n>m>1$, there does not exist $f$ as above.  The point is to consider the critical locus of the determinant.  The computation below proves that "cohomology with supports" has cohomological dimension equal to $2(n-1)$ for the pair of the quasi-affine scheme $\textbf{Mat}_{n\times n} \setminus \text{Crit}(\Delta_{n\times n})$ and its closed subset $\text{Zero}(\Delta_{n\times n})\setminus \text{Crit}(\Delta_{n\times n})$.  Since pushforward under affine morphisms preserve cohomology of quasi-coherent sheaves, that leads to a contradiction.   
Denote the $m\times m$ determinant polynomial by $$\Delta_{m\times m}:\textbf{Mat}_{m\times m} \to \mathbb{A}^1.$$  For integers $m$ and $n$ such that $m$ divides $n$, you ask whether there exists a homogeneous polynomial morphism $$f:\textbf{Mat}_{n\times n}\to \textbf{Mat}_{m\times m}$$ of degree $n/m$ such that $\Delta_{n\times n}$ equals $\Delta_{m\times m}\circ f$.  Of course that is true if $m$ equals $1$: just define $f$ to be $\text{Delta}_{n\times n}$.  Similarly, this is true if $m$ equals $n$: just define $f$ to be the identity.  Thus, assume that $2\leq m < n$; this manifests below through the fact that the critical locus of $\Delta_{m\times m}$ is nonempty.  By way of contradiction, assume that there exists $f$ with $\Delta_{n\times n}$ equal to $\Delta_{m\times m}\circ f$.
Lemma 1. The inverse image under $f$ of $\text{Zero}(\Delta_{m\times m})$ equals $\text{Zero}(\Delta_{n\times n})$.  In other words, the inverse image under $f$ of the locus of matrices with nullity $\geq 1$ equals the locus of matrices with nullity $\geq 1$.
Proof. This is immediate. QED
Lemma 2. The inverse image under $f$ of the critical locus of $\Delta_{m\times m}$ equals the critical locus of $\Delta_{n\times n}$.  In other words, the inverse image under $f$ of the locus of matrices with nullity $\geq 2$ equals the locus of matrices with nullity $\geq 2$.
Proof.
By the Chain Rule, $$d_A\Delta_{n\times n} = d_{f(A)}\Delta_{m\times m}\circ d_Af.$$  Thus, the critical locus of $\Delta_{n\times n}$ contains the inverse image under $f$ of the critical locus of $\Delta_{m\times m}$.  Since $m\geq 2$, in each case, the critical locus is the nonempty set of those matrices whose kernel has dimension $\geq 2$, this critical locus contains the origin, and this critical locus is irreducible of codimension $4$.  Thus, the inverse image under $f$ of the critical locus of $\Delta_{m\times m}$ is nonempty (it contains the origin) and has codimension $\leq 4$ (since $\textbf{Mat}_{m\times m}$ is smooth).  Since this is contained in the critical locus of $\Delta_{n\times n}$, and since the critical locus of $\Delta_{n\times n}$ is irreducible of codimension $4$, the inverse image of the critical locus of $\Delta_{m\times m}$ equals the critical locus of $\Delta_{n\times n}$. QED
Denote by $U_n\subset \text{Zero}(\Delta_{n\times n})$, resp. $U_m\subset \text{Zero}(\Delta_{m\times m})$, the open complement of the critical locus, i.e., the locus of matrices whose kernel has dimension precisely equal to $1$. By Lemma 1 and Lemma 2, $f$ restricts to an affine morphism $$f_U:U_n\to U_m.$$ 
Proposition. The cohomological dimension of sheaf cohomology for quasi-coherent sheaves on the quasi-affine scheme $U_n$ equals $2(n-1)$.
Proof.
The quasi-affine scheme $U_n$ admits a morphism, $$\pi_n:U_n \to \mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*,$$ sending every singular $n\times n$ matrix $A$ parameterized by $U_n$ to the ordered pair of the kernel of $A$ and the image of $A$.  The morphism $\pi_n$ is Zariski locally projection from a product, where the fiber is the affine group scheme $\textbf{GL}_{n-1}$.  In particular, since $\pi_n$ is affine,  the cohomological dimension of $U_n$ is no greater than the cohomological dimension of $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*$.  This equals the dimension $2(n-1)$.  
More precisely, $U_n$ is simultaneously a principal bundle for both group schemes over $\mathbb{P}^{n-1}\times(\mathbb{P}^{n-1})^*$ that are the pullbacks via the two projections of $\textbf{GL}$ of the tangent bundle.  Concretely, for a fixed $1$-dimension subspace $K$ of the $n$-dimensional vector space $V$ -- the kernel -- and for a fixed codimension $1$ subspace $I$ -- the image -- the set of invertible linear maps from $V/K$ to $I$ is simultaneously a principal bundle under precomposition by $\textbf{GL}(V/K)$ and a principal bundle under postcomposition by $\textbf{GL}(I)$.  In particular, the pushforward of the structure sheaf, $$\mathcal{E}_n:=(\pi_n)_*\mathcal{O}_{U_n},$$ is a quasi-coherent sheaf that has an induced action of each of these group schemes.  
The invariants for each of these actions is just $$\pi_n^\#:\mathcal{O}_{\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*}\to \mathcal{E}_n.$$  Concretely, the only functions on an algebraic group that are invariant under pullback by every element of the group are the constant functions.  The group schemes and the principal bundle are each Zariski locally trivial.  Consider the restriction of $\mathcal{E}_n$ on each open affine subset $U$ where the first group scheme is trivialized and the principal bundle is trivialized. The sections on this open affine give a $\mathcal{O}(U)$-linear representation of $\textbf{GL}_{n-1}$. Because $\textbf{GL}_{n-1}$ is linearly reductive, there is a unique splitting of this representation into its invariants, i.e., $\pi_n^\#\mathcal{O}(U)$, and a complementary representation (having trivial invariants and coinvariants).  The uniqueness guarantees that these splittings glue together as we vary the trivializing opens.  Thus, there is a splitting of $\pi_n^\#$ as a homomorphism of quasi-coherent sheaves, $$t_n:(\pi_n)_*\mathcal{O}_{U_n} \to \mathcal{O}_{\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*}.$$  
For every invertible sheaf $\mathcal{L}$ on $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*$, this splitting of $\mathcal{O}$-modules gives rise to a splitting, $$t_{n,\mathcal{L}}:(\pi_n)_*(\pi_n^*\mathcal{L})\to \mathcal{L}.$$  In particular, for every integer $q$, this gives rise to a surjective group homomorphism, $$H^q(t_{n,\mathcal{L}}):H^q(U_n,\pi_n^*\mathcal{L}) \to H^q(\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*,\mathcal{L}) .$$
Now let $\mathcal{L}$ be a dualizing invertible sheaf on $\mathbb{P}^{n-1}\times (\mathbb{P}^{n-1})^*.$  This has nonzero cohomology in degree $2(n-1)$.  Thus, the cohomological dimension of sheaf cohomology for quasi-coherent $\mathcal{O}_{U_n}$-modules also equals $2(n-1)$. QED 
Since $f_U$ is affine, the cohomological dimension for $U_n$ is no greater than the cohomological dimension for $U_m$.  However, by the proposition, the cohomological dimension for $U_m$ equals $2(m-1)$.  Since $1<m<n$, this is a contradiction.
A: 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.
A: This is not an answer, but it is to big to be a comment.
Let me write your matrix $X$ blockwise $(M_{\alpha\beta})_{1\le\alpha,\beta\le k}$, where the blocks are $k/d\times k/d$. If the blocks $M_{\alpha\beta}$ commutte to each other, then 
$$\det X=\det((d_{\alpha\beta})_{1\le\alpha,\beta\le k}$$
where $d_{\alpha\beta}:=\det M_{\alpha\beta}$.
A particular case of this property is the formula $\det(A\otimes B)=(\det A)^m(\det B)^n$ where $m$ is the size of $B$, $n$ that of $A$ (Thanks to Suvrit !).
As for the general case, I guess that the answer to your question is negative.
A: As others fascinatingly settled negative answer for the general question,
let me give a comment on certain situation when the answer is positive.
So if one has dxd matrix and d=k*n, we may split the matrix into nxn matrix which elements are kxk matrices. Now the statement is -  if these elements satisfy certain commutation relations (i.e. elements in each column commutate and cross-diagonal commutators are equal - see Manin matrix),
then $det_d = det_k det_n $ - i.e. first calculate determinant for nxn matrix whose elements are matrices - get a kxk matrix as an answer and then
calculate usual kxk determinant of the later.
(See section 5.5 page 36 here - sorry for self-advertising ). 
