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I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that:

Claim: Any homomorphism $\phi: {\mathbb Z}^{\infty} \to {\mathbb Z}$ that vanishes on the direct sum is identically zero.

Proof: (I learned this from a book by T.Y.Lam):

For a prime number $p$, let $A_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $(a_0, 2a_1pa_1, 4a_2,...)$p^2a_2,...)$, i.e. the elements whose$i$-th coordinate is divisible by$p^i$. Any element$x \in A_p$can be decomposed as$x= y+z= (a_0, 2a_1pa_1, \dots, 2^{n-1}a_{n-1}p^{n-1}a_{n-1}, 0,0, \dots ) + 2^n p^n (0,0,..,0, a_n, 2a_{n+1},..)$pa_{n+1},..)$

Now, $y$ is in the direct sum, hence $\phi(y)=0$. Also $\phi(z) \in 2^n p^n {\mathbb Z}$, which implies that $\phi (x) \in \cap_{n=1}^{\infty} 2^n p^n {\mathbb Z} = { 0 }$

Now, choose two distinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that $A_p+A_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.

3 added 390 characters in body

I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$in which . In the obvious quotient map, the image of the direct sum is mapped to zero. Now, it is enough to show that

claim: any

Claim: Any homomorphism from ${\mathbb Z}^{\infty}$ \phi: {\mathbb Z}^{\infty} \to ${\mathbb {\mathbb Z}$ that vanishes on the direct sum is identically zero.

Here is a proof for this

Proof: (I learned it this from a book by Lam)T.Y.Lam):

Since any element

For a prime number $p$, let $A_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $x= (a_0, (a_0, 2a_1, 4a_2,...)$, i.e. the elements whose $i$-th coordinate is divisible by $p^i$. Any element $x \in A_p$ can be expressed decomposed as

$x' x= y+z= (a_0, 2a_1, \dots, 2^{n-1}a_{n-1}, 0,0, \dots ) + 2^n (0,0,..,0, a_n, 2a_{n+1},..)$with

Now, $x'$ belonging to y$is in the direct sum, any such homomorphism sends such hence$x$to zero. The same argument works if you use powers of \phi(y)=0$. Also $3$. Now\phi(z) \in 2^n {\mathbb Z}$, since whichimplies that$gcd(2^n,3^n)=1$, any element \phi (x) \in the direct product is a sum of \cap_{n=1}^{\infty} 2^n {\mathbb Z} = { 0 }$

Now, choose two elements of these formsdistinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that$A_p+A_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.

I think the answer is no. The ultraproduct $U$ is a quotient of ${\mathbb Z}^{\infty}$ the direct product of countably many copies of ${\mathbb Z}$ in which the direct sum is mapped to zero. Now, it is enough to show that
claim: any homomorphism from ${\mathbb Z}^{\infty}$ to ${\mathbb Z}$ vanishes on the direct sum is identically zero.
Since any element of the form $x= (a_0, 2a_1, 4a_2,...)= \text{ an element in the direct sum} 4a_2,...)$ can be expressed as $x' + 2^n (0,0,..,0, a_n, 2a_{n+1},..)$ with $x'$ belonging to the direct sum, any such homomorphism sends such $x$ to zero. The same argument works if you use powers of $3$. Now, since $gcd(2^n,3^n)=1$, any element in the direct product is a sum of two elements of these forms.