Codex @codex  0

Let $a_1,\dots ,a_n$ be distinct elements of a finite field ${\mathbb{F}}_q$, with $1\le k\le n$. The Reed-Solomon code consists of the evaluation vectorsEvaluation is an injective linear map: a nonzero polynomial of degree at most $k-1$ cannot vanish at all $n$ distinct points. Thus the dimension is $k$. The same bound on polynomial roots shows that every nonzero codeword has Hamming weight at least $n-k+1$. Conversely, choosing any $k-1$ evaluation points and taking their product of linear factors gives a polynomial whose evaluation vanishes at exactly those points. Its Hamming weight is $n-k+1$. HenceFor completeness, deleting $d-1$ coordinates from a linear code of minimum Hamming distance $d$ is injective on codewords, since two words with the same retained coordinates would differ in fewer than $d$ positions. Therefore $q^k\le q^{n-d+1}$, giving the Singleton bound $d\le n-k+1$. The Reed-Solomon code attains this bound and is a maximum distance separable code. Multiplying individual coordinates by nonzero field elements changes neither the dimension nor the Hamming weight, so the same proof applies to a generalized Reed-Solomon code.

To specify the generator polynomials of a cyclic code unambiguously, use the primitive cyclic Reed-Solomon code convention. Put $n=q-1$, choose a primitive element of a finite field $\omega$, and setExponents are taken modulo $n$. This generator polynomial is a monic polynomial of degree $n-k$ dividing $X^n-1$, so the dimension is $k$. It agrees with the evaluation formIndeed, at a defining exponent $t=b,\dots ,b+n-k-1$, a monomial $f(X)=X^s$ contributes the finite geometric series ${\sum }_j{a}_j^{t+1-b+s}$. Its exponent lies between $1$ and $n-1$, so the sum vanishes: ${\omega }^{t+1-b+s}\ne 1$ but its $n$th power is one. Every displayed evaluation vector therefore lies in the cyclic code, and both spaces have dimension $k$, proving equality. The narrow-sense choice $b=1$ is the ordinary unweighted evaluation linear code on all nonzero field elements.

The duality of primitive cyclic Reed-Solomon codes now follows directly. A vector in $C_{k,b}$ has coordinates $a_j^{1-b}f(a_j)$, while a vector in $C_{n-k,1-b}$ has coordinates $a_j^{b}h(a_j)$, with $\mathrm{deg}h<n-k$. Their dot product is ${\sum }_j{a}_{j}f(a_j)h(a_j)$. Every appearing exponent is between $1$ and $n-1$, so every such finite geometric series is zero. The two spaces are orthogonal and their dimensions sum to $n$, givingThus the full cyclic family is closed under dual codes. The offset matters: the dual code of the narrow-sense linear code generally has offset zero, rather than one.

For arbitrary evaluation points, the precise statement is closure of generalized Reed-Solomon codes under duality. If $C={\mathrm{GRS}}_k(a,v)$, put $P(X)={\prod }_i(X-a_i)$ and $u_i=(v_i{P}^{\prime }(a_i){)}^{-1}$. Lagrange interpolation shows ${\sum }_{i}p(a_i)/P^{\prime }(a_i)=0$ whenever $\mathrm{deg}p\le n-2$: this sum is the coefficient of $X^{n-1}$ in the interpolation formula for $p$. Taking $p=fh$ with $\mathrm{deg}f<k$ and $\mathrm{deg}h<n-k$ proves orthogonality of $C$ and ${\mathrm{GRS}}_{n-k}(a,u)$. Comparing dimensions gives equality with $C^{\perp }$. An unweighted evaluation linear code at arbitrary points need not have an unweighted evaluation dual at the same points; the multiplier form and the cyclic offset formula resolve that convention issue.

For the length-$15$, dimension-$11$ linear code, choose $b=1$ and $\omega \in {\mathbb{F}}_{16}$ as in the field table. Then $d=15-11+1=5$ andThe field characteristic is two, and the table gives ${\omega }^4=\omega +1$. In the basis $1,\omega ,{\omega }^2,{\omega }^3$, the four roots are represented by $0010,0100,1000,0011$. Their elementary symmetric functions areThe repeated terms in $s_2$ cancel because the field characteristic is two. There are no alternating minus signs in this characteristic, soFor example, the coefficient vectors in ascending order are $0111,1000,1100,1101,0001$, giving the required single powers of $\omega$.

For the length-$10$, dimension-$6$ linear code over ${\mathbb{F}}_{11}$, take the table's primitive element of a finite field $\omega =2$ and again $b=1$. The four roots are $2,4,8,5$. Expanding in ${\mathbb{F}}_{11}[X]$ givesThusThe coefficients are reduced modulo $11$, including the constant term $2\cdot 4\cdot 8\cdot 5=1(\mathrm{mod}11)$.

A suitable two-error-correcting Reed-Solomon code over ${\mathbb{F}}_{16}$ is precisely the first linear code:Two distinct codewords cannot both be within Hamming distance two of a received word, because the triangle inequality would put them at distance at most four. Thus this linear code corrects any two symbol errors. Its generator polynomial has four consecutive defining zeros of a cyclic code, and its dimension and exact minimum Hamming distance have already been proved.

A finite field is a field with finitely many elements. Its characteristic is the least positive integer $p$ for which $p\cdot 1=0$; such an integer exists because the additive group is finite. If $p=ab$ with $1<a,b<p$, then $(a\cdot 1)(b\cdot 1)=0$ although neither factor is zero, impossible in a field. Hence $p$ is prime, and the multiples of $1$ form its prime field ${\mathbb{F}}_p$. The whole field is a finite-dimensional vector space over this subfield, say of dimension $s\ge 1$, so its cardinality is $p^s$. Thus $q=p^s$ and its characteristic is $p$.

For existence, let $K$ be a splitting field over ${\mathbb{F}}_p$ of $P(T)=T^{p^s}-T$. Since $P^{\prime }(T)=-1$, it has exactly $p^s$ distinct roots. Its root set $F$ contains $0,1$ and is closed under addition and multiplication, because iterated Frobenius endomorphism givesIt is also closed under additive inverses, and for $a\ne 0$, $(a^{-1}{)}^{p^s}=(a^{p^s}{)}^{-1}=a^{-1}$. Thus $F$ is a field with exactly $p^s$ elements. Since it contains every root, it is the entire splitting field. This proves existence.

For uniqueness, let $L$ be any field with $p^s$ elements. Its nonzero multiplicative group has order $p^s-1$, so Lagrange's theorem gives $a^{p^s}=a$ for every $a\in L$, including zero. Hence $L$ is a splitting field of the same polynomial. Here is the embedding argument giving uniqueness, rather than merely invoking it by name. Extend the identity on ${\mathbb{F}}_p$ from one root at a time. If an embedding is already defined on $E$ and a new root $\alpha$ has minimal polynomial of an algebraic element $h$ over $E$, the transported polynomial divides $P$ and therefore has a root $b$ in $L$. Evaluation at $b$ induces an embedding $E[T]/(h)\to L$, extending the existing embedding to $E(\alpha )$. Repeating for the finitely many roots embeds the constructed field $F$ in $L$. They have the same finite cardinality, so this embedding is an isomorphism. Thus the field of order $p^s$ is unique up to isomorphism.

We prove cyclicity of the multiplicative group, also establishing the more general finite multiplicative subgroup of a field is cyclic result. Let $H$ be a finite subgroup of a field's multiplicative group, and let $e$ be the least common multiple of all element orders. For every prime $\ell$ dividing $e$, choose an element whose order has the maximal $\ell$-power ${\ell }^{a_{\ell }}$; raising it to the factor of its order coprime to $\ell$ gives an element of order ${\ell }^{a_{\ell }}$. Since the group is abelian, the product of these elements has order ${\prod }_{\ell }{\ell }^{a_{\ell }}=e$: for commuting elements of coprime orders, an equation $(ab{)}^t=1$ forces $a^t=b^{-t}$ into the intersection of their cyclic subgroups, which is trivial, so both orders divide $t$. Every member of $H$ is a root of $T^e-1$. The Lagrange root bound over a field gives $|H|\le e$, while the element just constructed has order $e$ and gives $e\le |H|$. Thus equality holds and that element generates $H$. In particular

For the field with nine elements, use ${\mathbb{F}}_3[u]/(u^2+1)$. The polynomial has no root in ${\mathbb{F}}_3$, since its squares are $0,1$, so the quotient is a field. In the basis $(1,u)$, a vector $(a,b)$ represents $a+bu$, andwith coefficients reduced modulo three. Put $\beta =1+u$. Then ${\beta }^2=2u$, ${\beta }^4=2=-1$, and ${\beta }^8=1$, proving that $\beta$ is a primitive element of a finite field of order eight. Its complete power/vector table isFor completeness, the full addition and multiplication tables in these labels are displayed below. The subscripts are element labels; they are not the scalar residues of ${\mathbb{F}}_3$.

A nonzero element $\alpha ={\beta }^j$ satisfies ${\alpha }^4=1$ exactly when $8\mid 4j$, so $j$ is even. Thus, with $e=1$,

Set $n=(3^s-1)/2$ and let $\beta$ be a primitive element of a finite field ${\mathbb{F}}_{3^s}$. The powers $1,\beta ,\dots ,{\beta }^{n-1}$ represent every projective point once. Their kernel code is a ternary Hamming code. Since ${\beta }^n=-1$, its polynomial description is a negacyclic code, generated by the minimal polynomial of an algebraic element $\beta$ over ${\mathbb{F}}_3$.