Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 2 / b (Polynomial relations and Bézout coefficients)

The monomials $1,x,\dots ,x^n$ form a basis of $V_n$: they span every polynomial of degree of a polynomial at most $n$, and equality of polynomials forces equality of every coefficient, proving linear independence. ThereforeThe zero polynomial is included in $V_n$ even though its degree of a polynomial need not be assigned an ordinary nonnegative value.

Suppose, for a contradiction, that $p$ and $q$ have no common root of a polynomial. Since the polynomial ring $\mathbb{C}[x]$ has no zero divisors and $p,q\ne 0$, the relation $ap+bq=0$ with $a,b$ not both zero forces both $a$ and $b$ to be nonzero.

By the fundamental theorem of algebra, writewhere the ${\alpha }_j$ are distinct. For each $j$, $q({\alpha }_j)\ne 0$. The equality $bq=-ap$ implies that $(x-{\alpha }_j{)}^{m_j}$ divides $bq$. The multiplicity of a root in polynomial multiplication is the sum of its multiplicities in the factors: factor out the largest powers of $x-{\alpha }_j$ and note that the two remaining factors have nonzero values at ${\alpha }_j$. Since $q({\alpha }_j)\ne 0$, $b$ must contain $(x-{\alpha }_j{)}^{m_j}$.

Thus $b$ has at least ${\sum }_j{m}_j=n$ roots of a polynomial counted with multiplicity of a root, so $\mathrm{deg}b\ge n$. This contradicts $b\in V_{n-1}$. ConsequentlyCounting multiplicity of a root is necessary: counting only distinct roots of a polynomial would miss repeated factors.

Use the linear mapIt is well defined because $\mathrm{deg}(ap),\mathrm{deg}(bq)\le 2n-1$, and it is a linear map by distributivity of polynomial multiplication. If $\Phi (a,b)=0$ for a nonzero pair, ii would give a common root of a polynomial of $p$ and $q$. The hypothesis therefore implies $\ker \Phi =\{(0,0)\}$, so $\Phi$ is injective.

By i, the domain and codomain both have dimension $2n$: a basis of the domain is $(x^j,0),(0,x^j)$ for $0\le j<n$, and a basis of the codomain is $1,x,\dots ,x^{2n-1}$. The stated finite-dimensional result makes $\Phi$ surjective. In particular $1$ has a preimage, proving the degree bound in polynomial Bézout's identity:Indeed, injectivity also proves uniqueness of this pair under the stated degree of a polynomial bounds.