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No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, this should work now. Hopefully there are no other major errors].

This does get rather messy, so let's start simple and construct an example showing that being torsion is not a stalk-local property.

Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, it is possible to build up a counterexample bit-by-bit by starting with a ring satisfying some relatively simple conditions and progressively adding generators and relations to force the required properties to hold. This does get quite messy though and this answer went through a couple of revisions before I fixed up the construction, but it is still rather long (!!) (there is a possibility that there is already some well-known ring which happens to have the required properties, but I couldn't think of one). Actually, the construction below is not quite as terrifying as it might appear: the idea is simply to start with a polynomial ring in three indeterminates $x,y,z$, successively add elements to force the ideal generated by $x,y$ to consist of zero-divisors, then add some relations to force $x,y$ to be regular in the localizations at $z,1-z$ respectively. The difficult bit is showing that these steps do not interfere with each other by identifying properties which remain stable throughout the construction.

First, I'll introduce a bit of notation. I will work with (always commutative, unitial) algebras over the base (polynomial) ring $R=\mathbb{Z}[x,y,z]$. An R-algebra is then just a ring $A$ together with a homomorphism $R\to A$ or, simply, a ring together with three distinguished points $x,y,z\in A$. An R-morphism is then the same thing as a ring homomorphism respecting distinguished points.

Given an R-algebra $A$, let $K(A)\subseteq A$ denote the smallest ideal such that all $a\in A$ satisfy $ax\in K(A)\Rightarrow az\in K(A)$ and $ay\in K(A)\Rightarrow a(1-z)\in K(A)$. It can be seen that, if $K(A)=0$, then $x$ is a regular element in $A_z$ and $y$ is regular in $A_{1-z}$. We will be interested in the case $K(A)$ is a proper ideal, and will make use of the following property: $A$ satisfies property (P) if $K(A/(a))$ is proper for every $a\in Ax+Ay$. This property is preserved under R-morphisms.

(1) Let $f\colon A\to B$ be a morphism of R-algebras. Then $f(K(A))\subseteq K(B)$. Furthermore,

• If $I\subseteq A$, $J\subseteq B$ are ideals with $f(I)\subseteq J$ and $K(B/J)$ is proper, then $K(A/I)$ is proper.
• If $B$ satisfies (P) then so does $A$.

We can find R-algebras satisfying property (P) easy enough.

(2) If $A$ is a non-trivial commutative ring then the polynomial ring $R\otimes A\cong A[x,y,z]$ satisfies (P).

Case 2: Whenever $a(x_0,y_0,z_0)=0$ then $x_0y_0=0$. This means that $xy$ is contained in the radical ideal generated by $a$, so $a$ divides $x^ry^r$ some $r\ge1$. This means that $a$ is a multiple of $x$ or $y$ and one of $(x_0,y_0,z_0)=(0,1,0)$ or $(1,0,1)$ satisfies the required condition. So, $K(B/(a))$ is proper.

(3) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ with a left-inverse and such that, for every $a\in Ax+Ay$, there is a $b\in B$ with $ab=0$ and $K(B/(a,1-b))$ is proper.

The condition that $K(B/(a,1-b))$ is proper implies that $B/(a,1-b)$ is a non-trivial ring, so $b\not=0$. However, the property that $b\not=0$ will not be stable under the constructions I will use, which is why it is stated like this. Also, as $f$ has a left inverse, (1) says that $B$ satisfies (P).

Now, for a fixed $a\in I$, set $b=J+X_a$, so $ab=0$. Also, consider the morphism $A[(X_c)_{c\in I}]\to A\to A/(a)$ taking each $X_c$ to 0 (for $c\not=a$) and $X_a$ to 1. As its kernel contains $J$, it defines a morphism $g\colon B\to A/(a)$, which takes $a$ to 0 and $b$ to one. Therefore, the ideal $Ba+B(1-b)$ maps to 0 and, as $K(A/(a))$ is proper, (1) says that $K(B/(a,1-b))$ is proper.

We can piece these extensions together so that $Ax+Ay$ is forced to consist of zero divisors.

(4) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ such that, for every $a\in Bx+By$ there is a $b\in B$ with $ab=0$ and $K(B/(a,1-b))$ is proper.

Set $A_0=A$ and use (3) to construct a sequence of extensions $f_i\colon A_i\to A_{i+1}$ with left inverses such that, for every $a\in A_ix+A_iy$ there is a $b\in A_{i+1}$ with $ab=0$ and $K(A_{i+1}/(a,1-b))$ is proper.

Next we can force $x$ and $y$ to be units in $A_z$ and $A_{1-z}$ respectively.

(5) Suppose that $A$ satisfies the following: for every $a\in Ax+Ay$ there is a $b\in A$ with $ab=0$ and $K(A/(a,1-b))$ is proper. Then, the R-algebra $B=A/K(A)$ satisfies these same properties and also $K(B)=0$.

That $K(B)=0$ follows quickly from the definition of $K$. Suppose $a\in Ax+Ay,b\in A$ are such that $ab=0$ and $K(A/(a,1-b))$ is proper. Set $C=A/(a,1-b)$, so that $C/K(C)$ is not trivial. By (1), the canonical morphism $A\to C$ maps $K(A)$ into $K(C)$. So, it induces a morphism $B\to C/K(C)$. This takes $a$ and $1-b$ to zero, so it induces an R-morphism $B/(a,1-b)\to C/K(C)$. As $K(C/K(C))=0$, (1) implies that $K(B/(a,1-b))$ maps to zero, so is proper.

This construction has led to the promised counterexample.

For any $a\in I$ there is a $b\in B$ with $K(B/(a,1-b))$ proper. In particular, $(a,1-b)$ must be a proper ideal, so that $b\not=0$ and $a$ is a zero divisor.

No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, this should work now. Apologies for the length of this answer, but I don't see any quick constructions].

This does get rather involved, so let's start simple and construct an example showing that being torsion is not a stalk-local property.

Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, I will start with a simple example of a polynomial ring and then transform it in such a way as to give the properties we are looking for. I find it helpful to first fix the following notation: Start with the base (polynomial) ring $R=\mathbb{Z}[x,y,z]$. A (commutative, unitial) R-algebra is simply a ring with three distinguished elements $x,y,z$, and a morphism of R-algebras is just a ring homomorphism respecting these distinguished elements. For an R-algebra $A$, define $K(A)\subseteq A$ to be the smallest ideal such that, for all $a\in A$, $$ \begin{align} ax\in K(A)&\Rightarrow az\in K(A),\\\\ ay\in K(A)&\Rightarrow a(1-z)\in K(A). \end{align} $$ In particular, $K(A)=0$ implies that $x$ is a regular element in the localization $A_z$ and $y$ is a regular element in $A_{1-z}$. If we can construct such an example where the ideal $Ax+Ay$ consists purely of zero divisors, then that will give the counterexample needed. The idea is to start with $A=\mathbb{Z}[x,y,z]$ and transform it using the following steps.

• Force the elements of $I=Ax+Ay$ to be zero divisors. So, for each $a\in I$, add an element $b$ to $A$ in as free a way as possible such that $ab=0$. Adding elements to $A$ also has the effect of adding elements to $I$. So, this step needs to be iterated to force these new elements of $I$ to also be zero divisors.
• Replace $A$ by the quotient $A/K(A)$ to force the condition $K(A)=0$.

The first step above is easy enough. However, we do need to be careful to check that the second step does not undo the first. Suppose that $a\in A$ is a zero divisor, so that $ab=0$ for some non-zero $b$. It is possible that taking the quotient in the second step above takes $b$ to zero, so that $a$ becomes a regular element again. To get around this, we need some stronger condition on $b$ which implies $b\not=0$ and is also stable under taking the quotient. Note that $A(1-b)$ being a proper ideal or, equivalently, $A/(1-b)$ being nontrivial, will imply that $b\not=0$. In turn, this is implied by $K(A/(1-b))$ being a proper ideal. As it turns out, this property of $b$ does remain stable under each of the steps above, and can be used to show that this construction does give the counterexample required. However, note that if $ab=0$ and $K(A/(1-b))$ is proper, then $a=a(1-b)\in A(1-b)$, from which we can deduce that $K(A/(a))$ is a proper ideal. This necessary condition is unchanged by either of the steps above, so we had better check that elements $a\in Ax+Ay$ in our R-algebra do satisfy this from the outset. I'll make the following definition: $A$ satisfies property (P) if $K(A/(a))$ is proper for every $a\in Ax+Ay$. As it turns out, polynomial rings do satisfy this property and, consequently, the construction outlined above works fine.

Now on to the details of the argument.

(1) Let $f\colon A\to B$ be an R-morphism. Then $f(K(A))\subseteq K(B)$. Furthermore,

• If $I\subseteq A$, $J\subseteq B$ are ideals with $f(I)\subseteq J$ and $K(B/J)$ is proper, then $K(A/I)$ is proper.
• If $B$ satisfies (P) then so does $A$.

(2) If $A$ is a non-trivial ring then the polynomial ring $R\otimes A\cong A[x,y,z]$ satisfies (P).

Case 2: Whenever $a(x_0,y_0,z_0)=0$ then $x_0y_0=0$. This means that $xy$ is contained in the radical ideal generated by $a$, so $a$ divides $x^ry^r$ some $r\ge1$. Then $a$ is a multiple of $x$ or $y$ and one of $(x_0,y_0,z_0)=(0,1,0)$ or $(1,0,1)$ satisfies the required condition. So, $K(B/(a))$ is proper.

(3) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ with a left-inverse and such that, for every $a\in Ax+Ay$, there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

Now, for a fixed $a\in I$, set $b=J+X_a$, so $ab=0$. Consider the morphism $A[(X_c)_{c\in I}]\to A\to A/(a)$ taking each $X_c$ to 0 (for $c\not=a$) and $X_a$ to 1. As its kernel contains $J$, it defines a morphism $g\colon B\to A/(a)$, which takes $b$ to one. Therefore, the ideal $B(1-b)$ maps to 0 and, as $K(A/(a))$ is proper, (1) says that $K(B/(1-b))$ is proper.

(4) If $A$ satisfies (P), then we can construct an R-morphism $f\colon A\to B$ such that, for every $a\in Bx+By$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ is proper.

Set $A_0=A$ and use (3) to construct a sequence of extensions $f_i\colon A_i\to A_{i+1}$ with left inverses such that, for every $a\in A_ix+A_iy$ there is a $b\in A_{i+1}$ with $ab=0$ and $K(A_{i+1}/(1-b))$ is proper. Note that, as each $f_i$ has a left inverse, (1) says that $A_{i+1}$ satisfies (P) whenever $A_i$ does. So, we can keep applying (3) to build up the entire sequence of extensions.

(5) Suppose that $A$ satisfies the following: for every $a\in Ax+Ay$ there is a $b\in A$ with $ab=0$ and $K(A/(1-b))$ is proper. Then, the R-algebra $B=A/K(A)$ satisfies the same property, and also $K(B)=0$.

That $K(B)=0$ follows quickly from the definition of $K$. Suppose $a\in Ax+Ay,b\in A$ are such that $ab=0$ and $K(A/(1-b))$ is proper. Set $C=A/(1-b)$, so that $C/K(C)$ is not trivial. By (1), the canonical morphism $A\to C$ maps $K(A)$ into $K(C)$. So, it induces a morphism $B\to C/K(C)$. This takes $1-b$ to zero, so it induces an R-morphism $B/(1-b)\to C/K(C)$. As $K(C/K(C))=0$, (1) implies that $K(B/(1-b))$ maps to zero, so is proper.

For any $a\in I$ there is a $b\in B$ with $ab=0$ and $K(B/(1-b))$ proper. In particular, $(1-b)$ must be a proper ideal, so that $b\not=0$ and $a$ is a zero divisor.

Now set $B=k[x,y,z]$ and choose $a\in Bx+By$. The idea is to look at the morphism $\theta\colon B/(a)\to k$ taking $x,y,z$ to some $x_0,y_0,z_0\in k$ with $a(x_0,y_0,z_0)=0$. As long as these satisfy $ux_0\Rightarrow uz_0=0$ and $uy_0\Rightarrow u(1-z_0)=0$ (all $u\in k$) then $K(B/(a))$ will be contained in the kernel of $\theta$, so will be proper. For this to be the case it is enough that both ($x_0\not=0$ or $z_0=0$) and ($y_0\not=0$ or $z_0=1$).

Now set $B=k[x,y,z]$ and choose $a\in Bx+By$. The idea is to look at the morphism $\theta\colon B/(a)\to k$ taking $x,y,z$ to some $x_0,y_0,z_0\in k$ with $a(x_0,y_0,z_0)=0$. As long as these satisfy $ux_0=0\Rightarrow uz_0=0$ and $uy_0=0\Rightarrow u(1-z_0)=0$ (all $u\in k$) then $K(B/(a))$ will be contained in the kernel of $\theta$, so will be proper. For this to be the case it is enough that both ($x_0\not=0$ or $z_0=0$) and ($y_0\not=0$ or $z_0=1$).

Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, it is possible to build up a counterexample bit-by-bit by starting with a ring satisfying some relatively simple conditions and progressively adding generators and relations to force the required properties to hold. This does get quite messy though and this answer went through a couple of revisions before I fixed up the construction, but it is still rather long (!!) (there is a possibility that there is already some well-known ring which happens to have the required properties, but I couldn't think of one).

Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, it is possible to build up a counterexample bit-by-bit by starting with a ring satisfying some relatively simple conditions and progressively adding generators and relations to force the required properties to hold. This does get quite messy though and this answer went through a couple of revisions before I fixed up the construction, but it is still rather long (!!) (there is a possibility that there is already some well-known ring which happens to have the required properties, but I couldn't think of one). Actually, the construction below is not quite as terrifying as it might appear: the idea is simply to start with a polynomial ring in three indeterminates $x,y,z$, successively add elements to force the ideal generated by $x,y$ to consist of zero-divisors, then add some relations to force $x,y$ to be regular in the localizations at $z,1-z$ respectively. The difficult bit is showing that these steps do not interfere with each other by identifying properties which remain stable throughout the construction.