I'm looking for some feedback on a construction I came across. Loosely, it entails sending a vector space isomorphically to a vector space of polynomials, 'restoring' the ring structure, and asking whether the polynomial is reducible. I want to claim that the polynomial in question is irreducible if and only if the vector to which it corresponds is irreducible over the tensor product. I am having difficulty specifically with the 'restoration' of the ring structure. It isn't obvious to me that there is a natural way to do this and I haven't seen anything in the several books on algebra I've looked into.
Here is the formal construction:
Let $\mathcal{H}$ be a finite-dimensional complex vector space of dimension $d$. Let the basis of this space be labeled $\{|q\rangle\}$ with $q \in \{0,...,d-1\}$. Consider the $n^{th}$ tensor power of $\mathcal{H}$, $T^n(\mathcal{H}) = \mathcal{H} \otimes ... \otimes \mathcal{H}$ $n$ times.
Let
$$ |\phi\rangle = \sum_{q_1,...,q_n = 0}^{d-1} C_{q_1,...,q_n} |q_1\rangle \otimes ... \otimes |q_n\rangle $$
and
$$ \phi(\vec{z}) = \sum_{q_1,...,q_n = 0}^{d-1} C_{q_1,...,q_n} z_1^{q_1}...z_n^{q_n} $$
Consider the following map:
$$f: |\phi\rangle \mapsto \phi(\vec{z})$$
This is a linear isomorphism between vector spaces $f: T^n(\mathcal{H}) \rightarrow \Phi_{d-1}[\mathbb{C};z_1,...,z_n]$ where $\Phi_{d-1}[\mathbb{C};z_1,...,z_n]$ is the complex vector space of polynomials of degree $d-1$ in each of the $n$ indeterminates. I hope this is clear, but I can supply the proof if necessary. (Note: the usual notation is to use the total degree as the index of the vector space, but in this case that would be $(d-1)^n$ and I want to consider on the subspace of $\Phi_{(d-1)^n}[\mathbb{C}]$ for which each indeterminate has degree at most $d-1$ per the map above; that is, I want to exclude basis elements like $z_k^{(d-1)^n}$ for example.)
Now I can regard the image of $f$ as a subset of the ring of polynomials in $n$ indeterminates. If we do this, then we can naturally ask whether $\phi$ is reducible. The problem I have is this: I don't see a natural way to do this. Nevertheless, all of the examples I've considered so far satisfy the claim. In fact, they satisfy the stronger claim that the degree of reducibility of the polynomial is equivalent to the degree of reducibility of the vector (by degree of reducibility I just mean the integer which corresponds to the number of irreducible factors--disregarding the unit--in the decomposition of the polynomial/vector).
Does anyone see a way to formalize this construction? Or can anyone give a counterexample in which the polynomial is reducible but the corresponding vector is not?
Thanks for any help you can offer. Cheers.
As an interesting aside, this seems like a special case of a more general isomorphism I've come across in the field of quantum information. There we replace linear operators which act 'locally' by the operators indexed by an integer which indicates which 'component' of the tensor product we act on. For example, if we consider $\mathcal{H} \otimes \mathcal{H}$, a local unitary operator is of the form $U \otimes 1$, we can replace this by $U_1$ to indicate that the action of $U$ is on the 'first' vector space in the tensor product. Naively, it looks like we are restoring a sort of commutativity to the algebra of (local) operators when we do this, since if $U$, $V$ are linear operators acting on $\mathcal{H}$ then $U \otimes V = U_1 V_2 = V_2 U_1$. (Of course, in this case that is just saying $U \otimes V = (U \otimes 1)(1 \otimes V) = (1 \otimes V)(U \otimes 1)$.)
Edit:
I think it might be helpful to clarify with an example since it is possible that I am using incorrect terminology. Let $dim[\mathcal{H}] = 3$ and $n = 2$. Then we are considering the case that $\mathcal{H} = sp\{|0\rangle, |1\rangle, |2\rangle\}$ and the tensor power is $T^2(\mathcal{H}) = \mathcal{H} \otimes \mathcal{H}$. The isomorphism I refer to then maps the basis elements as follows:
$$|q_1 q_2\rangle \mapsto x^{q_1}y^{q_2}$$
The number of basis elements on the right hand side of this map is $3 \cdot 3 = 9$ and is a closed subspace of the usual vector space of polynomials of degree $4$. These spaces have the same dimension and using the map I defined above it is easy to show that this is linear and a bijection. I hope this clarifies my meaning.
Edit 2:
I think I should also carry the above example forward to further elucidate my meaning. I want to stress that the heart of my question gets at the formalization of the following, so it is possible that it is incorrect in its phrasing. I put quotes around words I know have a standard technical meaning which I intend to employ in an intuitive manner.
If we still work with $d = 3$ and $n = 2$, then we can compute: for any $|\phi_1\rangle, |\phi_2\rangle \in \mathcal{H}$, if $|\phi\rangle = |\phi_1\rangle \otimes |\phi_2\rangle$, then
$$ |\phi\rangle = \sum_{q_1, q_2 = 0}^{2}c_{q_1} c'_{q_2} |q_1\rangle \otimes |q_2\rangle \mapsto \phi(x,y) = \sum_{q_1, q_2 = 0}^{2}c_{q_1} c'_{q_2} x^{q_1}y^{q_2} \mapsto \left(\sum_{q_1}c_{q_1}x^{q_1}\right)\cdot\left(\sum_{q_2}c'_{q_2}y^{q_2}\right)$$
where this last map follows from the 'restoring' of the ring structure and the factorization follows from the fact that the composition of maps essentially gives
$$|\phi_1\rangle \otimes |\phi_2\rangle \mapsto \phi_1(x) \cdot \phi_2(y)$$
so that every vector factorable over $\otimes$ corresponds to a reducible polynomial in the codomain. In this case, the map appears as something like an algebra homomorphism. (Maybe this is the correct way to phrase the map to begin with?)
As a second, more concrete example, one can begin with an irreducible polynomial and invert to obtain an irreducible tensor product of vectors. For example, consider the polynomial
$$\phi(x,y) = c - c'xy$$
with $c, c' \in \mathbb{C}$. This is irreducible for all non-zero values of $c$, $c'$ in the ring of polynomials over $\mathbb{C}$ in two indeterminates. If we invert this, we obtain the vector
$$c - c'xy \mapsto c|0\rangle \otimes |0\rangle - c'|1\rangle \otimes |1\rangle$$
which is an irreducible vector over the tensor product. It is easy to extend this sort of calculation to higher dimension and larger number of indeterminates.
Edit: I misunderstood the question. Here is the phenomenon you appear to be interested in: the polynomial ring $k[x_1, \dots x_n]$ in $n$ variables can canonically be written as a tensor product
$$k[x_1] \otimes \dots \otimes k[x_n] \cong k[x]^{\otimes n}$$
of $n$ copies of the polynomial ring in one variable. The isomorphism does the obvious thing, sending $x_1^{k_1} \otimes x_2^{k_2}$ to the ordinary product $x_1^{k_1} x_2^{k_2}$ and so forth. If you restrict degrees you get finite-dimensional vector spaces. The one-variable polynomial $x^q$ corresponds to your $| q \rangle$.
And pure tensors correspond to polynomials which admit a "separation of variables" in the sense that they can be written as a product $f_1(x_2) \dots f_n(x_n)$. This is a different concept of reducibility / irreducibility than the usual concept for polynomials; e.g. the polynomial $x^2 - y^2 = (x - y)(x + y)$ is reducible as a polynomial but it does not admit a separation of variables.