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Suppose we have a (commutative) ring $R$ and an $R$-algebra $S$. Furthermore, suppose that $S\cong R^n$ as $R$-modules, that is, $S$ is free of rank $n$ as an $R$-module. Can we always choose $1$ to be a basis element for $S$? Equivalently, is it necessary that $S/R \cong R^{n-1}$?

If not, how about in the case that $R=\mathbb{Z}$?

This is true in the case $n=1$: if we have a ring homomorphism $\phi: R\to S$ and an $R$-module isomorphism $\psi: S\to R^1$, it's not hard to show that $\phi$ must also be an isomorphism, making $S/R\cong R^0$.

And it is true if $n=2$ and $R=\mathbb{Z}$: in that case it is known that $S\cong \mathbb{Z}[x]/p(x)$, where $p$ is some degree 2 monic polynomial, so that in particular $S$ is generated freely as a $\mathbb{Z}$-module by 1 and $x$.

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@Graham L.: This is not finitely generated. I think that Owen wants $S$ to be finitely generated free as a module over $R$. – Martin Brandenburg Feb 20 '12 at 20:23
Perhaps we should also assume $S \neq 0$; otherwise $S=0$ and $n=0$ is a counterexample. – Martin Brandenburg Feb 20 '12 at 20:27
up vote 9 down vote accepted

This is not true in general. For example, assume that $P$ is a projective module on $R$ that is not free, but such that $P \oplus R$ is free (there are many such examples). Set $S= R \oplus P$, and give $S$ an $R$-algebra structure by taking the products of two elements of $P$ to be 0.

On the other hand, it is true when $R$ is a Dedekind domain (so, in particular, for $R = \mathbb Z$). Let us show that $S/R$ is torsion-free. If not, there would be some $s \notin R$ and $a, b \in R$, $a ≠ 0$, such that $as = b$. On the other hand $S$ is integral over $R$, so there is a relation $s^n + a_1 s^{n-1} + \dots + a_{n-1}s + a_n = 0$, with all the $a_i$ in $R$. Multiplying this by $a^n$ gets you a relation in $R$; by dividing this by $a^n$ in the quotient field $K$ of $R$ shows that $r := b/a \in K$ is integral over $R$. Since $R$ is normal we have $r\in R$, and $as = ar$. Since multiplication by $a$ is injective in $S$ we see that $s =r$ is in $R$, and this is a contradiction.

So $S/R$ is torsion free and finite; hence it is projective, and $S$ is isomorphic to $R \oplus (S/R)$ as an $R$-module. But over a Dedekind domain every stably free module is free, and this concludes the proof.

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Thanks so much! Just what I was looking for. – Owen Biesel Feb 20 '12 at 21:18

The following is a counterexample to the question you asked but not to a variant you might have asked. Let $M$ be a projective $R$-module which is not free but such that $R\oplus M\cong R^n$ (any module which is stably free but not free gives rise to such an $M$). Let $S=R\oplus M$ with ring structure given by $M^2=0$. Then $S\cong R^n$ as an $R$-module, but $S/R\cong M$ is not free.

A different question you might ask is not for $S/R$ to be free but for it to be projective. This is equivalent to asking $S$ to split as a direct sum $R\oplus S/R$. Since projective is equivalent to locally free, this is the same as asking your original question with the additional assumption that $R$ is a local ring. I don't know the answer in this case, though I would guess that there's a clever way to build a counterexample.

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As others have mentioned, your claim isn't necessarily true. It seems to be true though that we may write $S=1\cdot R \oplus P$ for some projective $R$-module $P$. This shows that $1_S$ can be choosen as a basis element for instance if $R=K[x_1,\ldots,x_n]$ or if $R$ is a PID (basically whenever there is a reason for a projective module to be free).

Let $X_1,\ldots,X_n$ be an $R$-basis of $S$, and write $$ S \cong R[x_1,\ldots,x_s]/(x_ix_j -\sum_{k}r_{ijk} x_k,\ a_1x_1+\ldots+a_nx_n-1) $$ such that sending $X_i$ to $x_i$ is an isomorphism. I claim the ideal $(a_1,\ldots, a_n)_R$ in equal to $(1)_R$. If that is so, then there are $t_1,\ldots,t_n\in R$ such that $\sum t_i a_i = 1$, and therefore $$\varphi: \ S\longrightarrow R: X_i \mapsto t_i$$ is an $R$-module homomorphism mapping $1_S$ to $1_R$. Sending $1_R$ to $1_S$ defines a splitting of this epimorphism, and therefore $S \cong 1_S\cdot R \oplus {\rm ker}(\varphi)$. So $S$ is a direct sum of $R$ and the projective $R$-module ${\rm ker}(\varphi)$.

All that is left to show is that $(a_1,\ldots,a_n)_R = (1)_R$ actually holds. Assume this is not true, and suppose $a_1,\ldots,a_n$ are all contained in some maximal ideal $\mathfrak m$ of $R$. Then, by looking at the relation $a_1 x_1+\ldots+a_nx_n-1=0$ mod $\mathfrak m$, we see that $S/\mathfrak m S =0$, which is a contradiction, since $S$ is assumed to be free of rank $n$ over $R$.

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For the first question, it is nevertheless true that if $S$ is an $R$-algebra (containing $1$) that is finitely generated and projective as an $R$-module, then $S \simeq R \oplus S/R$ as $R$-modules, i.e., $R$ is a direct summand of $S$. See Lemma 1.3 of (I'm sure there are other references, but I couldn't find one easily when I was writing this.)

I guess it is clear by now that the "equivalently" part is not true.

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Thanks! At some point I started thinking of finite projective algebras as not just locally free, but locally admitting a basis that contains 1. The proof I eventually came across was that since $R\hookrightarrow S$ is a pure submodule and $S$ is flat, the quotient $S/R$ is also flat. It's also finitely presented if $S$ is, hence projective, hence the exact sequence $0\to R\to S\to S/R \to 0$ splits. – Owen Biesel Dec 5 '15 at 8:29

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