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I changed “epimorphisms” to “surjective ring maps”. The notion of a ring epimorphism is more general than that of a surjective ring homomorphism (and includes for instance localizations). It is clear from context that the poster specifically meant surjective ring map, not merely epimorphism.
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edit approved Jan 21, 2019 at 16:47
Neil Epstein
edit approved Jan 21, 2019 at 16:47
Neil Epstein
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Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct epimorphismssurjective ring maps $f_1,f_2,f_3$ onto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a product of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct epimorphisms $f_1,f_2,f_3$ onto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a product of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct surjective ring maps $f_1,f_2,f_3$ onto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a product of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

correct typos
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edit approved Jan 19, 2019 at 11:23
Steve Jacob
edit approved Jan 19, 2019 at 11:23
Steve Jacob
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Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct eprimorphismepimorphisms $f_1,f_2,f_3$ intoonto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a porudctproduct of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct eprimorphism $f_1,f_2,f_3$ into $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a porudct of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct epimorphisms $f_1,f_2,f_3$ onto $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a product of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

added 9 characters in body
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Uriya First
Uriya First
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Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$$$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct eprimorphism $f_1,f_2,f_3$ into $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a porudct of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ eprimorphism $f_1,f_2,f_3$ into $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a porudct of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

Consider the following subrings of $\mathbb{F}_2^3$: $$ V_1=\{(0,0,0),(1,1,1),(1,0,0),(0,1,1)\}\\ V_2=\{(0,0,0),(1,1,1),(0,1,0),(1,0,1)\}\\ V_3=\{(0,0,0),(1,1,1),(0,0,1),(1,1,0)\} $$ Then $\mathbb{F}_2^3=V_1\cup V_1\cup V_3$.

Let $R$ be a PID admitting $3$ distinct eprimorphism $f_1,f_2,f_3$ into $\mathbb{F}_2$ and write $$ R_i=\{a\in R: (f_1(a),f_2(a),f_3(a))\in V_i\} $$ for $i=1,2,3$. Then $R_1,R_2,R_3$ are subrings of $R$ with $R=R_1\cup R_2\cup R_3$. They are proper subrings by the Chinese Remainder Theorem.

To construct such $R$, take a cubic extension $K$ of $\mathbb{Q}$ in which $2$ splits into a porudct of $3$ primes $P_1,P_2,P_3$, and choose $R$ to be the localization of $\mathcal{O}_K$ at the multiplicative set $R\setminus(P_1\cup P_2\cup P_3)$.

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Uriya First
Uriya First
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