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I don't quite follow what you mean by "such that every block of $Q$ is some block of $D$ that one of its elements is removed." Can you rephrase it or explain exactly what you mean by an example? Did you mean $Q$ is a triple system instead of block size $4$? Or is it of order $v-1$ rather than $v$?
@Felix Ha ha. Looks like I beat you to it! This kind of construction is nicely explained in a very accessible way in "Design Theory" by Lindner and Rodger too if you're interested: books.google.com/books/about/…
@JAS The simplest way is to see if your code is linear or not. If it's not linear, it can't be a Reed-Solomon code. If it's linear, the weight distributions are all the same across (linear) MDS codes for a given length, dimension, and alphabet size. So, I doubt there is an extremely elementary way to distinguish two MDS codes, although I could be wrong.
Instead, you first construct a cyclic code from the generator polynomial $\prod_{i=1,2,\dots,2^{r-1}}(x-\alpha^i)$ and show that the columns of its parity-check matrix are the all $2^r-1$ vectors. If we prove that $\prod_{i=1,2,\dots,2^{r-1}}(x-\alpha^i)$ always defines a cyclic code and always results in such $H$, we're done: we just proved that binary Hamming codes are always cyclic.
Also, if you want a simple proof that all binary Hamming codes are cyclic, the proof given in this lecture note (Theorem 7.7) may be a good one. It's exactly the same as what I wrote in the answer except that this one only considers the binary case and the terminology is slightly more elementary. The point is that you don't try to re-arrange columns of $H$ that consists of all $2^r-1$ vectors of $\mathbb{F}_2^r\setminus\{\boldsymbol{0}\}$ into some cyclic form.
From now on, to avoid confusion, we write $x$ as $\alpha$ when talking about $\mathbb{F}_{16}$ (because cyclic codes also use polynomials and we want to use the symbol $x$ for generator polynomials and whatnot). So, if you reduce everything by $\alpha^4 = \alpha+1$, by expanding $(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)$ as $(x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)=(x-\alpha)(x-\alpha^2)(x-\alpha-1)(x-(\alpha+1)^2)=...$, you arrive at $x^4+x+1$.
An example of minimal polynomials of degree $4$ is $x^4+x+1$. If you pick this guy, computations over the extended field is modulo $x^4+x+1$, which means $x^4+x+1=0$ or equivalently $x^4=x+1$ because coefficients are from $\mathbb{F}_2$. You can show that $x$ is a primitive $15$th root of unity by directly checking $x^i$ are all different for $i=0,1,\dots,14$. Note that because our computation rule is "modulo chosen minimal polynomial," you use the relation $x^4 = x+1$ whenever you get $x^i$ with $i\geq 4$ so that everything is contained in the set $\{0,1,x,x+1,\dots,x^3+x^2+x+1\}$.
