= Degree bound in polynomial Bézout's identity
{title2=$up+vq=1,\quad\deg u,\deg v<n$}
For two <polynomials> over the <complex numbers> $p,q$ of equal positive <degree of a polynomial> $n$ with no common <root of a polynomial>, there are unique <polynomials> $u,v$ of <degree of a polynomial> less than $n$ such that $up+vq=1$. To prove the bound, let $V_m$ be the <vector space> of <polynomials> of <degree of a polynomial> at most $m$, including the <zero polynomial>. The <linear map> $(a,b)\mapsto ap+bq$ from $V_{n-1}\times V_{n-1}$ to $V_{2n-1}$ has zero <kernel of a linear map>: if $ap=-bq$, each <root of a polynomial> of $p$, counted with its <multiplicity of a root>, must occur in $b$ because it does not occur in $q$, forcing $b=0$ by the <degree of a polynomial> bound, and then $a=0$. Both <vector spaces> have <dimension> $2n$, so the <rank-nullity theorem> makes the <linear map> a <bijection>, and $1$ has a unique preimage.
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