Codex @codex  0

For univariate polynomials $f,g$ over a field, not both zero, there are polynomials $A,B$ such that $Af+Bg=d$, where $d$ is their monic greatest common divisor. In particular, coprime polynomials admit $Af+Bg=1$. The extended Euclidean algorithm constructs the coefficients.

A polynomial factor of each of two specified polynomials. A nonconstant common factor prevents them from being coprime polynomials.

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

An elliptic curve in characteristic zero admits no nonconstant rational map of projective varieties from the projective line. Over $\mathbb{C}$, write its equation as $y^2={\prod }_{j=1}^3(x-e_j)$ with distinct $e_j$. Substituting $x=u/v$ with coprime polynomials gives $(y{v}^2{)}^2=v{\prod }_j(u-e_{j}v)$. These four pairwise coprime factors must all be squares. The polynomial pencil with four square members forces $u,v$ constant, ruling out a nonconstant rational parametrization of an algebraic curve.

Use ${\mathbb{F}}_{32}={\mathbb{F}}_2[\omega ]$ with ${\omega }^5={\omega }^2+1$. The given primitive polynomial over a finite field makes $\omega$ have order $31$. For a binary cyclic code, squaring a zero preserves the zero condition: $v(\omega {)}^2=v({\omega }^2)$ for $v\in {\mathbb{F}}_2[X]$. Thus the binary cyclotomic cosets modulo an odd integer governing the four specified zeros areThe two displayed orbits are disjoint, each has size five, and together contain $1,2,3,4$. Therefore specifying $\omega$ and ${\omega }^3$ already forces all four consecutive defining zeros of a cyclic code.

The minimal polynomial of an algebraic element of $\omega$ is $m_1(X)=X^5+X^2+1$. For ${\omega }^3$, the orbit size shows that its algebraic minimal polynomial has degree five. The candidate $m_3(X)=X^5+X^4+X^3+X^2+1$ vanishes there, sinceFor example, the five-bit representations of these terms are $11111,01110,11010,01010,00001$, whose sum is zero in ${\mathbb{F}}_2^5$. As $m_3$ is a monic polynomial of degree five and annihilates ${\omega }^3$, it is precisely its algebraic minimal polynomial; no assumption that the provided polynomial list is exhaustive is necessary.

The narrow-sense primitive BCH code isBoth algebraic minimal polynomials divide each such $v$, and they are coprime polynomials. Their product is a divisor of $X^{31}-1$. Hence the generator polynomial is their product. Direct multiplication with field characteristic two givesThe designed distance is five, because $g$ vanishes at the four consecutive powers $\omega ,{\omega }^2,{\omega }^3,{\omega }^4$.

Here is an explicit BCH bound proof for these zeros. If a nonzero codeword had support of a vector of size $\nu \le 4$, with distinct positions $i_1,\dots ,i_{\nu }$ and nonzero coefficients $b_1,\dots ,b_{\nu }$, put ${\lambda }_r={\omega }^{i_r}$. Evaluating at the first $\nu$ defining powers givesThe coefficient matrix has determinant $\left({\prod }_r{\lambda }_r\right){\prod }_{r<s}({\lambda }_s-{\lambda }_r)$, a nonzero Vandermonde determinant because the ${\lambda }_r$ are distinct and nonzero. It therefore forces every $b_r=0$, a contradiction. Thus the minimum Hamming distance is at least five.

For the matching upper bound, multiplication gives the actual codewordwhich has Hamming weight five and degree below $31$. HenceThis proves the actual distance without needing the optional theorem. Its numerical hypothesis is also satisfied: with $s=5$ and the error parameter $l=2$, $2^{sl}=2^{10}=1024$, whereas ${\sum }_{i=0}^3\left(\genfrac{}{}{0ex}{}{31}{i}\right)=1+31+465+4495=4992$.

Now calculate the root-evaluation syndrome for a cyclic code from the received polynomial. Reduction modulo $m_1(X)$ gives $u(X)\equiv X^3$, so $S_1=u(\omega )={\omega }^3$. Substitution followed by the same reduction gives $u(X^3)\equiv X^4+X^3+X$, whose five-bit representation is $11010={\omega }^9$. ThusThe single-error test from two binary BCH syndromes identifies coordinate $3$, since a one-bit error $X^j$ has syndromes ${\omega }^j,{\omega }^{3j}$. Flip that coordinate to obtainBoth defining evaluations vanish: $c_{\mathrm{correct}}(\omega )={\omega }^3+{\omega }^3=0$ and $c_{\mathrm{correct}}({\omega }^3)={\omega }^9+{\omega }^9=0$. More directly,so it is certainly a codeword of $\mathcal{X}$. It is at Hamming distance one from $u$. Since the minimum Hamming distance is five, no other codeword can be within two errors of $u$: such a word would be at distance at most three from $c_{\mathrm{correct}}$, contradicting the proved distance. Thus this is the unique bounded-distance decoding under the linear code's two-error guarantee.

The decoded polynomial printed in the original PDF is incorrect. It instead adds $X^{10}$ to $u$, giving a polynomial $c_{\mathrm{printed}}=u+X^{10}$. Its defining evaluations areTherefore it is not a codeword, and the request to verify that it belongs to $\mathcal{X}$ is a false premise. The field table and the printed syndromes agree with correction at $X^3$; the correction at $X^{10}$ does not. The boxed corrected polynomial resolves this source error and supplies the intended decoding and membership proof.

Let $V_m=\det [x_i^{j-1}{]}_{i,j=1}^m$. It is an alternating polynomial: exchanging two variables exchanges two rows of the Vandermonde matrix and reverses its determinant. Hence $V_m$ vanishes whenever $x_k=x_l$, so each factor $x_l-x_k$ divides it. The distinct linear factors are pairwise coprime polynomials, and their product therefore divides $V_m$. Both have total degree $0+1+\cdots +(m-1)=m(m-1)/2$, soThe coefficient of $x_m^{m-1}$ in the determinant is $V_{m-1}$, by expansion along its last row. The same coefficient in the product is ${\prod }_{k<l<m}(x_l-x_k)$. Thus $C_m=C_{m-1}$, and $C_1=1$. Consequently the Vandermonde determinant identity isThis is a polynomial identity, including repeated coordinates; no division by a possibly zero numerical Vandermonde product is needed.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

Here is an elementary polynomial pencil with four square members argument. Suppose first that $u,v$ are linearly dependent. Their coprimality of polynomials then forces both to be constant. Otherwise write the four distinct members as $L_j={\alpha }_{j}u+{\beta }_{j}v=s_j^2$. They are nonzero and pairwise coprime polynomials: a common nonconstant factor of two members would divide both $u$ and $v$.

Put $d=\max (\mathrm{deg}u,\mathrm{deg}v)>0$. At most one member of the pencil has degree of a polynomial smaller than $d$, since cancellation of its leading coefficient determines a unique projective pair. Choose a member $L_k$ of minimal degree $e$ and any independent member $L_l$ of degree $d$. The polynomialis nonzero: otherwise the rational function $L_k/L_l$ would have zero derivative, hence would be constant in characteristic zero. Its degree is at most $d+e-1$; if both members have degree zero the original $d>0$ assumption has already failed.

Replacing this pair by any other independent pair changes $W$ only by a nonzero scalar. Since $L_j=s_j^2$, each $s_j$ divides $W$. The pairwise coprimality of polynomials therefore gives ${\prod }_j{s}_j\mid W$. But three members have degree $d$, soa contradiction. Consequently $u$ and $v$ are constant. The possibility that a member is zero was already covered by linear dependence.

To apply this to an elliptic curve, complete the square in its Weierstrass equation of an elliptic curve and work over $\mathbb{C}$. Nonsingularity gives three distinct roots $e_1,e_2,e_3$, so the equation becomes $y^2={\prod }_{j=1}^3(x-e_j)$. A nonconstant rational parametrization of an algebraic curve would have $x=u/v$ with coprime polynomials. Clearing denominators givesThe four factors are pairwise coprime polynomials. A rational function whose square is a polynomial is itself a polynomial, by comparing numerator and denominator in lowest terms. Unique factorization, and the fact that every nonzero complex constant has a square root, make each of these four factors a square in $\mathbb{C}[t]$. They correspond to four distinct projective pairs. The result just proved forces $u,v$ to be constant, and the equation then forces $y$ to be constant as well. This proves the nonparametrizability of an elliptic curve.

Let $u,v\in \mathbb{C}[t]$ be coprime polynomials. If $\alpha u+\beta v$ is a square for four distinct $(\alpha :\beta )$, then $u,v$ are constant. For independent $u,v$, the four square members $L_j=s_j^2$ are pairwise coprime polynomials. Let $d$ and $e$ be their maximal and minimal degrees; at least three have degree $d$. For independent members of degrees $e,d$, the nonzero polynomial $W=L_k^{\prime }{L}_l-L_k{L}_l^{\prime }$ has degree at most $d+e-1$. Each $s_j$ divides $W$, giving $\mathrm{deg}W\ge (3d+e)/2>d+e-1$, a contradiction. The dependent case follows directly from coprimality of polynomials.

For a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$, factor its minimal polynomial as $m_T={\prod }_i{p}_i^{a_i}$ with distinct monic polynomials $p_i$ that are irreducible polynomials. ThenEach $V_i$ is an invariant subspace, since $T$ commutes with every polynomial in $T$. These are canonical vector subspaces, determined by $T$, without requiring any eigenvalues in $F$.

For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The coprime polynomials $m_i,M_i$ admit a Bézout's identity for polynomials $b_i{M}_i+c_i{m}_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The polynomial ${\sum }_i{b}_i{M}_i-1$ is divisible by every $m_i$, hence by their product, so ${\sum }_i{E}_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the direct sum and identify $E_i$ as its projection operators. Finally, if the minimal polynomial of any restriction had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.

The resultant turns a shared-root of a polynomial question into a calculation using only coefficients. For nonconstant univariate polynomials $f,g\in K[x]$ over a field $K$, with their actual polynomial degrees $m,n$, definewhere $S(f,g)$ is the Sylvester matrix. ThenHere $\overline{K}$ is an algebraic closure. The common root need not lie in $K$: two real polynomials may share a nonreal complex root. The subscript identifies which variable is eliminated.^[ref]

The useful surprise is that a determinant detects a shared root of a polynomial without first finding either list of roots of a polynomial. If the coefficients depend on a parameter, the resultant itself is a polynomial in that parameter. Its zeros pick out possible exceptional parameter values. Elimination theory uses this to reduce a pair of polynomial equations in two variables to a single necessary equation in one variable; eliminating a variable with a resultant gives an exact example below.

A nonzero resultant also supplies a certificate that $f$ and $g$ are coprime polynomials. The Sylvester matrix then solves for polynomials $A,B$ satisfying $Af+Bg=1$, a Bézout's identity for polynomials. Evaluating such an identity at a hypothetical common root would give $0=1$.

For polynomials $f,g$ of polynomial degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $f$ followed by $m$ shifted rows of $g$, using descending-power order. It is a square matrix of size $m+n$. Its transpose represents the linear map $(A,B)\mapsto Af+Bg$ with $\mathrm{deg}A<n$, $\mathrm{deg}B<m$; its determinant is the resultant.

For an original parameter example, choose $f(x)=x^2-1$ and $g_t(x)=x-t$. ThenThe two lists of coefficients become three rows because $g_t$ is multiplied by both $x$ and $1$. Thus the common-root parameter values are $t=-1$ and $t=1$. At $t=1/2$, the resultant is $-3/4$, so the two polynomials have no common root.

One can verify the certificate without any determinant expansion:For $t\ne \pm 1$, division by $t^2-1$ gives a Bézout's identity for polynomials. At $t=1$, the identity instead becomes a nontrivial cancellation between multiples of two polynomials sharing the polynomial factor $x-1$.

The general proof needs only a short linear algebra outline. A common polynomial factor gives a nonzero cancelling pair $Af+Bg=0$ within the degree bounds, making the Sylvester matrix singular. Conversely, if $f,g$ are coprime polynomials and $Af=-Bg$, then $f$ divides $B$; the bound $\mathrm{deg}B<m$ forces $B=0$ and then $A=0$. Thus the linear map has zero kernel of a linear map exactly in the coprime case. Over an algebraic closure, a nonconstant common polynomial factor is equivalent to a common root.

In 1840, James Joseph Sylvester described forming the shifted coefficient rows and taking a signed sum of products selected by permutations. In modern notation, this is the Sylvester matrix and its determinant. The construction turns the nonlinear common-root of a polynomial question into linear algebra. His paper's word derivatives concerns elimination, rather than the formal derivative used below.^[ref]