Let $k$ be a field and $A$ a graded $k$-algebra with $A_0=k$, $A^i=0$ for $i<0$ and $\dim_kA^i<\infty$. Let $B$ be a subalgebra of $A$ which is normal, i.e. $AB^{\geq1}=B^{\geq1}A$. Assume that there is a finite subset $E$ of $A$ such that $E$ is a basis of $A$ as a graded right $B$-module. Do you know how to prove that $E$ is also a basis of $A$ as a graded left $B$-module?
Why is a left basis also a right basis?
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abstract-algebra
noncommutative-algebra
graded-rings
graded-modules
1 Answers
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Let $F$ be the free graded left $B$-module on $E$; there is a unique homomorphism of graded left $B$-modules $f:F\to A$ that is the identity on $E$. Note that $F^i$ and $A^i$ have the same (finite) dimension for all $i$, since $A$ is freely generated by $E$ as a graded right $B$-module (and so in particular, $A$ is isomorphic as a graded vector space to a direct sum of shifted copies of $B$ indexed by $E$). Since a linear map between finite-dimensional vector spaces of the same dimension is injective iff it is surjective, it suffices to show that $f$ is surjective. That is, it suffices to show $E$ generates $A$ as a left $B$-module.
To prove this, we use the fact that $E$ generates $A$ as a right $B$-module and normality. Let $a\in A$ be a homogeneous element; we will show by induction on the degree of $a$ that $a$ is in the left $B$-module generated by $E$. Since $E$ generates $A$ as a right $B$-module, we can write $$a=\sum_{e\in E}eb_e$$ for some homogeneous elements $b_e\in B$ with $|b_e|+|e|=|a|$ for each $e$. We will show each term $eb_e$ is in the left $B$-module generated by $E$.
If $|b_e|=0$, then $b_e\in k$ so $eb_e=b_ee$ is in the left $B$-module generated by $E$.
If $|b_e|\neq 0$, then $b_e\in B^{\geq 1}$, so $eb_e\in AB^{\geq 1}=B^{\geq 1}A$. Thus we can write $$eb_e=\sum b_na_n$$ where each $b_n\in B^{\geq 1}$ and $|a_n|+|b_n|=|eb_e|=|a|$ for each $n$. In particular, $|a_n|<|a|$ for each $n$, so by induction we know $a_n$ is in the left $B$-module generated by $E$. Thus $eb_e=\sum b_na_n$ is also in the left $B$-module generated by $E$.
Thus every term $eb_e$ in our expression for $a$ is in the left $B$-module generated by $E$, and hence so is $a$.