Codex @codex  0

Work over the finite field ${\mathbb{F}}_2$. Reduce polynomials modulo $X_i^2-X_i$, so their monomials are square-free. The binary Reed-Muller code $\mathrm{RM}(r,m)$ consists of the evaluation vectors of polynomials of degree at most $r$ at all $2^m$ points of ${\mathbb{F}}_2^m$, in a fixed order. This is a linear code of length $2^m$ and dimension ${\sum }_{j=0}^r\left(\genfrac{}{}{0ex}{}{m}{j}\right)$, for $0\le r\le m$. Square-free monomials give independent evaluation functions, as follows inductively by splitting according to the last variable.

The minimum Hamming distance of a linear code equals its least nonzero Hamming weight. The result isHere $r=0$ is the repetition code, while $r=m$ is the full binary vector space and has distance one. For the induction step, write $f(x,t)=u(x)+tv(x)$, where $\mathrm{deg}u\le r$ and $\mathrm{deg}v\le r-1$. Its evaluation vector is $(u,u+v)$, the Reed-Muller bar-product recursion. If $v=0$, its nonzero Hamming weight is at least $2d(r,m-1)=2^{m-r}$, with the full-space endpoint treated as above. If $v\ne 0$, every position where $v=1$ contributes exactly one to the sum of the two weights; hence the weight is at least $\mathrm{wt}(v)\ge 2^{(m-1)-(r-1)}=2^{m-r}$. The monomial $X_1\cdots X_r$ is one exactly on $2^{m-r}$ points, attaining the bound. Degrees greater than $m$ give the same full code; the zero code at negative degree has no nonzero word whose weight could define a finite minimum.

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.

For length-$n$ binary linear codes $C_2\subseteq C_1$, their bar product of binary linear codes isThe map $(u,v)\mapsto (u,u+v)$ is injective and linear, soIf $d_i$ is the minimum distance of $C_i$, thenFor $v\ne 0$ this is at least $d_2$; for $v=0$ a nonzero word has weight at least $2d_1$. Words $(0,v)$ and $(u,u)$ attain these bounds, proving

A parity-check matrix $P$ for a code $C$ is a matrix satisfying $C=\ker P$. A pair $(x,y)$ belongs to $C_1|C_2$ exactly whenThus a parity-check matrix isThis is the parity-check matrix of a bar product.

The Reed-Muller code $\mathrm{RM}(d,r)$ is the length-$2^d$ binary code obtained by evaluating every square-free polynomial in $d$ Boolean variables of degree at most $r$ on all points of ${\mathbb{F}}_2^d$. Equivalently,The square-free monomials form a basis, soA word in $\mathrm{RM}(d,1)$ evaluates an affine function $a_0+a\cdot x$. The two constant functions have weights $0$ and $2^d$. Every nonconstant affine function takes each value equally often, so all other codewords have weight $2^{d-1}$.

A monomial of degree $j$ has weight $2^{d-j}$. Hence every generator, and therefore every codeword, has even weight when $r<d$. For $r=d$, the monomial $x_1\cdots x_d$ has weight one. All words have even weight exactly for $0\le r\le d-1$.