Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 36 (Algebraic Coding)

For a linear code $X\le {\mathbb{F}}^n$, its dual code isThe standard bilinear form is nondegenerate. The map ${\mathbb{F}}^n\to X^{\ast }$ obtained by restricting the dot-product functional to $X$ is surjective: extend a functional from a basis of $X$ to a basis of ${\mathbb{F}}^n$, then use the nondegenerate dot product to represent it. Its kernel is $X^{\perp }$. The rank-nullity theorem therefore gives

The first proposed description of the dual code is false without self-orthogonality. At length three take $X=⟨(1,0,0)⟩$. It has the specified dimension, butwhereas $⟨X,(1,1,1)⟩$ contains $(1,0,0)$ and $(1,1,1)$, neither of which is in $X^{\perp }$. The natural corrected statement is the odd-length binary self-orthogonal dual extension: if also $X\subseteq X^{\perp }$, every word of $X$ has even Hamming weight, the all-ones word lies in $X^{\perp }$ but not in $X$, and dimensions then give $X^{\perp }=X\oplus ⟨1⟩$.

For a binary self-dual code, $x\cdot x=0$ for every $x\in X$. In ${\mathbb{F}}_2$, this equals ${\sum }_i{x}_i$, so every codeword has even Hamming weight. Thus $1\cdot x=0$ for all $x$, giving $1\in X^{\perp }=X$. The dimension identity yields $2\dim X=n$, so $n$ is even. Alternatively $1\in X$ and $1\cdot 1=0$ itself gives evenness. Hence the second assertion is true.

For every $n=2m$, useThis linear code has dimension $m$, and $(u,u)\cdot (v,v)=2(u\cdot v)=0$. Therefore $X\subseteq X^{\perp }$; both have dimension $m$, so equality holds. This constructs a binary self-dual code at every even length. Conversely the dimension identity already proves every self-dual length-$n$ linear code has dimension $n/2$.

For a nonbinary example, over ${\mathbb{F}}_5$ take $X=⟨(1,2)⟩$. Its generator has self-product $1+2^2=0$ in ${\mathbb{F}}_5$, so $X\subseteq X^{\perp }$. Both are one-dimensional, proving

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 $e=1$,

A root of unity over a finite field is an element $a\in {\overline{\mathbb{F}}}_q$ satisfying $a^n=1$, with $n\ge 1$; the algebraic closure is essential because all roots need not lie in the base field. Write $p=\mathrm{char}{\mathbb{F}}_q$ and $n=p^{a}m$, where $p\nmid m$. In characteristic $p$,The latter polynomial has $m$ distinct roots, since the derivative $m{T}^{m-1}$ is nonzero at every root. Those roots are closed under multiplication and inversion and contain $1$, so $E(n,q)$ is a finite multiplicative subgroup of a field. The proof in the preceding solution makes it a cyclic group. In particularThis also explains why the coprimality qualification is needed.

The multiplicative group of a finite field ${\mathbb{F}}_{q^s}^{\times }$ is cyclic of order $q^s-1$. It contains the entire group of $m$th roots precisely when $m\mid q^s-1$. For necessity, a generator of $E(n,q)$ has order $m$ and Lagrange's theorem applies. For sufficiency, a cyclic group whose order is divisible by $m$ contains $m$ roots of $T^m-1$; since there are exactly $m$ roots in the algebraic closure, these are all of them. Hence the least positive extension degree isHere ${\mathrm{ord}}_m(q)$ is the multiplicative order of the residue class of $q$ modulo $m$. If $n$ is coprime to $q$, replace $m$ by $n$. Existence of this order follows because $q$ is a unit modulo $m$.

A primitive $n$th root of unity over a finite field has multiplicative order exactly $n$. If $p\mid n$, no such element exists, since all roots have order dividing $m<n$. If $\gcd (n,q)=1$, choose a generator $\zeta$ of $E(n,q)$. The element ${\zeta }^j$ has order $n/\gcd (n,j)$, so it is primitive precisely when $\gcd (n,j)=1$. Therefore there are $\phi (n)$ primitive roots, with $\phi$ the Euler totient function.

If $\omega$ is primitive of order $n$, then $\omega \in {\mathbb{F}}_{q^{\ell }}$ exactly when ${\omega }^{q^{\ell }}=\omega$, equivalently $n\mid q^{\ell }-1$. The implication from that equality to membership follows because the roots of $T^{q^{\ell }}-T$ form ${\mathbb{F}}_{q^{\ell }}$, as proved in the preceding solution. Consequentlyfor $n>1$, and the degree is one for $n=1$. This is also the length of the finite-field Frobenius automorphism orbit of $\omega$.

Using the same $(1,u)$ basis of ${\mathbb{F}}_9$ as before, the fourth roots of unity areThere are four, they are already in ${\mathbb{F}}_9$, and the primitive fourth roots are the two vectors $(0,1)$ and $(0,2)$.

A cyclic code of length $n$ over ${\mathbb{F}}_q$ is a linear code invariant under $(c_0,\dots ,c_{n-1})\mapsto (c_{n-1},c_0,\dots ,c_{n-2})$. Associate the word with $c(T)={\sum }_{i=0}^{n-1}{c}_i{T}^i$. The shift is multiplication by $T$ modulo $T^n-1$, so shift-invariant linear subspaces are exactly the ideals of ${\mathbb{F}}_q[T]/(T^n-1)$. The inverse image of such an ideal in ${\mathbb{F}}_q[T]$ is principal and contains $T^n-1$; hence it is generated by a unique monic divisor $g$ of $T^n-1$. The code has basis $g,Tg,\dots ,T^{n-\mathrm{deg}g-1}g$, and dimension $n-\mathrm{deg}g$.

The defining zeros of a cyclic code are the roots of its generator $g$ in a splitting field. Since $g\mid T^n-1$, each is a root of unity over a finite field. Every codeword vanishes at each such root. If the characteristic divides $n$, root multiplicities must additionally be retained; in the Hamming construction below the modulus has no repeated roots.

The unrestricted ternary cyclic-equivalence assertion is false. At $s=2$ it would give a cyclic $[4,2,3{]}_3$ code. Butover ${\mathbb{F}}_3$, with the quadratic irreducible. A dimension-two cyclic code needs a degree-two generator, so its only possibilities are $T^2-1$ and $T^2+1$. Each is a weight-two codeword. Therefore neither has minimum distance three, and no code equivalent to the ternary $[4,2,3]$ Hamming code can be cyclic. A valid general version of the asserted construction assumes odd $s$. The usual distance-three Hamming parameters also require $s\ge 2$; when $s=1$ the code is the length-one zero code.

Here is the construction with its precise qualification. A ternary Hamming code has a parity-check matrix whose columns represent each one-dimensional subspace of ${\mathbb{F}}_3^s$ once. There are $n=(3^s-1)/2$ such subspaces. Its rank is $s$, because the coordinate lines are included. Its minimum distance is at least three, because no column is zero and no two are proportional. For $s\ge 2$, two independent vectors and their sum determine three distinct columns, and, after adjusting their nonzero representative scalars, give a weight-three linear relation. Thus its parameters are $[n,n-s,3{]}_3$. Permuting columns and changing their nonzero representatives produces a monomial equivalence of linear codes.

Take odd $s\ge 3$ and choose $\alpha \in {\mathbb{F}}_{3^s}^{\times }$ of order $n$. Since $n$ is odd, $-1$ is not in the cyclic subgroup generated by $\alpha$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ lie in distinct one-dimensional ${\mathbb{F}}_3$-subspaces: a proportional pair would have ratio $1$ or $-1$, neither possible for two distinct powers in this list. By their number they represent all projective points. Consequentlyis monomially equivalent to the ternary Hamming code. Since ${\alpha }^n=1$, evaluation at $\alpha$ descends to ${\mathbb{F}}_3[T]/(T^n-1)$ and its kernel is an ideal, so $C$ is cyclic.

The degree of $\alpha$ over ${\mathbb{F}}_3$ is $s$. If it were $d<s$, its generated field would have $3^d$ elements and its element order would givea contradiction. Its minimal polynomial of an algebraic element therefore has degree $s$. Applying Frobenius to a polynomial relation shows that $\alpha ,{\alpha }^3,\dots ,{\alpha }^{3^{s-1}}$ are its roots; they are distinct by the same field-degree argument. Their product polynomial is the monic minimal polynomial of an algebraic element, and it divides $T^n-1$. ThusThis identifies all the defining zeros. For $s=3$, $n=13$, the defining exponents are $1,3,9$ modulo $13$.

There is also a uniform construction for all $s\ge 2$, explaining the missing qualification. Let $\beta$ generate ${\mathbb{F}}_{3^s}^{\times }$, which has order $2n$. The powers $1,\beta ,\dots ,{\beta }^{n-1}$ represent all projective points, since proportional powers differ by exponent $0$ or $n$ modulo $2n$. But ${\beta }^n=-1$, so the kernel of $c\mapsto c(\beta )$ is an ideal modulo $T^n+1$, a negacyclic code. Multiplication by $T$ gives $(-c_{n-1},c_0,\dots ,c_{n-2})$, preserving its kernel. This code is always equivalent to the ternary Hamming code. If $s$ is odd, $n$ is odd too, and the substitution $T\mapsto -T$, or coordinate scaling $c_i\mapsto (-1{)}^i{c}_i$, changes this negacyclic construction into the cyclic one. In fact $\alpha =-\beta$ then has order $n$. These explicit constructions prove the qualified assertion rather than relying on a theorem name.

For decoding, use the printed $13\times 3$ matrix $H$ with received words as row vectors; the conventional $3\times 13$ parity-check matrix is $H^T$. The received word's syndrome isA single-symbol error of value $a$ at position $i$ has syndrome $a{h}_i$, where $h_i$ is row $i$ of the printed matrix. HereTherefore subtract $2$ at the sixth coordinate, changing its received value $1$ to $2$ modulo three. The decoded codeword isIts syndrome is zero. The thirteen projective rows and the two nonzero scalars give all $26$ nonzero syndromes exactly once, so this is the unique codeword at distance one. It is the transmitted word under the usual assumption of at most one symbol error. If positions $1,2,5$ are designated check positions, their independent unit rows make the other ten symbols systematic data, namely $(2,0,2,0,0,2,1,1,2,0)$.