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There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every localization of $M$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

$\def\mm{\mathfrak{m}}$ So we now assume that $R$ is local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every $M_{\mathfrak{p}}$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

$\def\mm{\mathfrak{m}}$ So we now assume that $R$ is local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every localization of $M$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

So we now assume that $R$ be local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

$\def\mm{\mathfrak{m}}$ Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every localization of $M$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

$\def\mm{\mathfrak{m}}$ So we now assume that $R$ is local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

There are no examples with $R$ local and $M$ finitely generated. I think you should be able to reduce the general case to the local case, and you can certainly do so when $R$ is noetherian, but I have a nagging suspicion that the property "torsion-free" does not localize as well as I want it to for non-noetherian rings. I don't have much intuition for the non-finitely-generated case.

$\def\mm{\mathfrak{m}}$ Proof: Suppose that $M$ is not flat. Let $R$ be local, $\mm$ the maximal ideal and $k = R/\mm$. Let $V = M \otimes \mm$. Let $f_i$ be a basis of $V$, for $i=1$ to $n$, and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.

There are no examples with $M$ finitely generated. We first reduce to the case that $M$ is local. Suppose that $M^{\otimes n}$ is torsion-free for all $n$. Then any localization of $M^{\otimes n}$ is torsion-free. (See, for example, Exercise 14.5.I in Ravi's notes. I had a nagging suspicion that there was a noetherian hypothesis needed for this, but Ravi is pretty careful about that and he doesn't give one.) So, if we have proved the local case, then we know that every localization of $M$ is flat. Flatness can be checked locally; see Ravi's proposition 25.2.3.

So we now assume that $R$ be local, with $\mm$ the maximal ideal and $k = R/\mm$. Suppose that $M$ is not flat. Let $V = M \otimes k$ and let $n = \dim_k V$. We will show that $M^{\otimes n}$ has torsion.

$\def\mm{\mathfrak{m}}$ Proof: Let $f_i$ be a basis of $V$ and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.

Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.

Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.

Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.