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I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.

An attempt at redemption: I actually found a reference on when the dual of direct product of a ring is direct sum:

www-users.mat.uni.torun.pl/~gregbob/seminars/2008.11.07b.pdf

I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.

An attempt at redemption: I actually found a reference on when the dual of direct product of a ring is direct sum:

http://www-users.mat.umk.pl/~gregbob/seminars/2008.11.07b.pdf

I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.

I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.

An attempt at redemption: I actually found a reference on when the dual of direct product of a ring is direct sum:

www-users.mat.uni.torun.pl/~gregbob/seminars/2008.11.07b.pdf

I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.</>

I think so. Let $f$ be a homomorphism from $\mathbb Z[[x]] $ to $\mathbb Z$. WRONG: Let $f(x^i)=a_i$. Then since $f(1+x+x^2...) \in \mathbb Z$, we must have $a_i=0$ for $i\gg 0$. So each map can be identified with an element in $\mathbb Z[x]$.