$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Spec{Spec}$Let $A$ be a commutative ring not necessarily with unit and $\mathbb{Z}_2 =\{0,1\}$ be the field of two elements. I am looking for a paper describing the structure of the set of ring homomorphisms $\Hom(A,\mathbb{Z}_2)$. So this is a specific subset of the set of all homomorphisms $A\to\mathbb{Z}_2$ as abelian groups.

If $\phi:A \to \mathbb{Z}_2$ is a non-zero ring homomorphism, then its kernel is a two-sided maximal ideal of $A$ having index $2$ as an abelian subgroup of $A$. And conversely, every such ideal $K$ determines a unique ring homomorphism $A \to A/K = \mathbb{Z}_2$.

Notice that $K$ can be regarded as a certain point of the spectrum $\Spec(A)$. Hence $\Hom(A,\mathbb{Z}_2)$ is a subset of $\Spec(A)$.

In the paper

  • Joseph A. Gallian, James Van Buskirk. The number of homomorphisms from $\mathbb{Z}_m$ into $\mathbb{Z}_n$, Amer. Math. Monthly 91 (1984), no. 3, 196-197, doi:10.1080/00029890.1984.11971375

it is proved that the number of ring homomorphisms $\Hom(\mathbb{Z}_m,\mathbb{Z}_n)$ equals $2^{\omega(n)-\omega(n/\gcd(m,n))}$, where $\omega(a)$ is the number of distinct prime factors of $a$. See also this question on MathOverflow.

In particular,

  • $\omega(1)=0$,
  • $\omega(2)=1$,
  • $\Hom(\mathbb{Z}_{2m},\mathbb{Z}_2)$ consists of two homomorphisms(zero and $x\mapsto x~\mathrm{mod}~2$),
  • $\Hom(\mathbb{Z}_{2m+1},\mathbb{Z}_2)$ consists of a unique zero homomorphism.

Could you please provide any references to the papers studying the set of ring homomorphisms $\Hom(A,\mathbb{Z}_2)$.

  • $\begingroup$ Are you sure Gallian and Buskirk are looking at ring homomorphisms? Or are they counting arbitrary additive homomorphisms? $\endgroup$ Oct 23, 2022 at 7:22
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    $\begingroup$ Yes, they even wonder that they did not find such a description in the literature before. Please look at their paper here: https://www.d.umn.edu/~jgallian/homs.pdf. They compute both types. An interesting notice is that the number of group homomorphisms $\mathbb{Z}_m\to\mathbb{Z}_n$ equals $gcd(m,n)$ and it is symmetric in $m$ and $n$, while the number of ring homomorphisms is not. $\endgroup$ Oct 23, 2022 at 7:30
  • $\begingroup$ The Gallian and Buskirk calculation includes homomorphisms which do not preserve 1. In your case, only the zero homomorphism is multiplicative and does not preserve 1: you should exclude this case from your count. $\endgroup$ Oct 23, 2022 at 7:42
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    $\begingroup$ The only ring homomorphsm $A\to \mathbb Z_2$ that does not preserve 1 is the zero homomorphism. $\endgroup$ Oct 23, 2022 at 7:51
  • 9
    $\begingroup$ Let $J$ be the intersection of all kernels of homomorphisms $A\to\mathbf{F}_2$ and $B=A/J$. Then $B$ is the largest Boolean algebra quotient of $A$. The set you're interested is the spectrum of $B$. It has a natural Hausdorff topology (compact totally disconnected), which coincides with the Zariski topology of $\mathrm{Spec}(B)$. If $A$ is noetherian, then so is $B$, and hence $B$ is finite and $\mathrm{Spec}(B)$ has cardinal $\log_2(|B|)$. (The above concerns unital homomorphisms, but as already mentioned, including non-unital homomorphisms just adds the zero homomorphism.) $\endgroup$
    – YCor
    Oct 23, 2022 at 9:05


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