Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 26 / 1 / ii / Solution

Start from the integral triangular basis of level-one modular forms. Proceed downwards in $j$. Set $g_{d-1}=F_{d-1}$; once $g_{j+1},\dots ,g_{d-1}$ are constructed, subtract from $F_j$ its coefficients at $q^{j+1},\dots ,q^{d-1}$ times these respective forms. The already constructed rows have a single nonzero coefficient among those initial positions, so this removes every unwanted initial coefficient without changing the coefficient one at $q^j$. All subtractions use integers. The resulting integral echelon basis of level-one modular forms therefore hasIts first $d$ coefficient vectors are the standard coordinate vectors, hence it is a basis. If another form had the same first $d$ coefficients as $g_j$, their difference would have those coefficients all zero; the triangular basis proves that difference vanishes. This proves uniqueness of every $g_j$, not merely uniqueness up to basis change.

For any $f\in M_k$, its expansion in this basis is $f={\sum }_{j=0}^{d-1}{a}_j(f)g_j$, since its first $d$ coefficients determine the basis coordinates. Comparing subsequent coefficients givesThe coefficients of $f$ need not be integral for this last identity. When needed at earlier indices, extend the notation by $c_n(j)={\delta }_{nj}$ for $0\le n<d$. This agrees with the actual coefficients of the basis forms and is the convention needed for the next part's all-index wording.