# What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?

That is, we take the set of set of all functions $\mathbb N \to \mathbb Z$ with componentwise operations to make it a ring with 1, and we want maps to $\mathbb Z$ preserving all operations (addition, multiplication, and 1).

Clearly the projection onto each factor $\mathbb Z$ is a homomorphism. What others are there?

Bonus question: what can we say more generally about $R$-algebra homomorphisms $R^\mathbb{N} \to R$, for a commutative ring $R$?

This is closely related to my earlier question Ideals of infinite product rings, still unanswered. I'm hoping this one will be easier.

• I think they will all be of the type $f(x_1,\dots)=\sum_i a_i x_i$ for some $a_i$, almost all of them $0$. Commented Jan 11, 2015 at 20:39
• I think that we need more information. Are both multiplication and addition componentwise? You allow infinitely many nonzero entries, I guess, so that the multiplicative unit would be $1$ everywhere? Commented Jan 11, 2015 at 20:45
• @Myself Those are indeed the only group homomorphisms - so automatically (ignoring the unit, which some folks care about ) they're the only ring homomorphisms.
– user98602
Commented Jan 11, 2015 at 20:59
• @MikeMiller: Even if we don't require a ring homomorphism to preserve the unit, that only gives us the zero homomorphism. Except for that, since $\mathbb Z^{\mathbb N}$ happens to contain a unit, that unit needs to map to $1$ in $\mathbb Z$, or the whole thing wouldn't be a homomorphism. Commented Jan 11, 2015 at 21:05
• @Ivan, his ring is not $\oplus\Bbb Z$. Commented Jan 12, 2015 at 3:53

Those are the only homomorphisms. (Except, if you don't require ring homomorphisms to preserve $1$, for the constant zero morphism).

Let $e_i$ be the sequence whose $i$th element is $1$ and the rest is $0$.

If $f:\mathbb Z^{\mathbb N}\to\mathbb Z$ is a ring homomorphism, then at most one of $f(e_0), f(e_1), \ldots, f(e_n), \ldots$ can be nonzero. Namely, if $i\ne j$, then $e_ie_j=0$ and therefore $f(e_ie_j)=f(e_i)f(e_j)=0$, so one of $f(e_i)$ and $f(e_j)$ must be zero.

Thus $f$ agrees with some multiple of a projection on the ideal generated by the $e_i$s. Without loss of generality, let's assume that $f(x_0,x_1,\ldots)=ax_0$ for some $a$ whenever cofinitely many $x_i$s are zero.

The following argument is adapted from an MO post that Mike Miller pointed to in a comment.

Choose sequences $(\alpha_i)_i$, $(\beta_i)_i$ such that $2^i\mid \alpha_i$ and $3^i\mid \beta_i$ and $\alpha_i+\beta_i=1$. (This is possible due to Bézout's identity since $2^i$ and $3^i$ are always coprime).

Now to show $f(x_0,x_1,x_2,\ldots)=ax_0$ in general write $$f(x_0,x_1,x_2,x_3,\ldots_)= \underbrace{f(x_0,0,0,\ldots)}_{ax_0} +\underbrace{f(0,\alpha_1 x_1,\alpha_2 x_2,\alpha_3 x_3,\ldots)}_P +\underbrace{f(0,\beta_1 x_1, \beta_2 x_2, \beta_3 x_3, \ldots)}_Q$$ I claim that each of the two last terms is zero. For any $n$ we have $$P = \underbrace{f(0,\alpha_1 x_1, \ldots, \alpha_{n-1} x_{n-1}, 0, 0, \ldots)}_0 + \underbrace{f(0,0,\ldots,0,\alpha_n x_n,\alpha_{n+1} x_{n+1}, \ldots)} _{\text{divisible by }2^n}$$ The only integer that is divisible by $2^n$ for all $n$ is zero, so $P$ must be $0$. Similarly $Q$ must be $0$.

So $f(x_0,\ldots)=ax_0$, and since $f(1,1,1,\ldots)=1$ (otherwise $f$ is not a ring homomorphism) we get $a=1$ and $f$ is a projection.

Bonus answer: This argument works whenever $R$ is a UFD with at least two different prime elements.

• Same argument repeated again and again. Commented Jan 11, 2015 at 22:25
• When you want to conclude $f(e_i)f(e_j) = 0$ isn't it simpler to use $e_ie_j = 0$ and the fact that $f$ is a ring homomorphism? Commented Jan 11, 2015 at 23:32
• @Joachim: Yes, much. Will edit, thanks. Commented Jan 11, 2015 at 23:37