Skip to main content
MathOverflow

Return to Answer

added 104 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$ in $R$. The map that sends an element of $R$ to its $C$-coset representative in $L''$ is the retraction.)

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$.)

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$ in $R$. The map that sends an element of $R$ to its $C$-coset representative in $L''$ is the retraction.)

added 38 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a transversalcomplete irredundant set of coset representatives for $C$.)

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a transversal for $C$.)

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$.)

added 189 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any tuplepair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a transversal for $C$.)

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any tuple in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$.

  • $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors. This means that the composition of the embedding $R\leq S\times S$ with either projection $\pi_i\colon S\times S\to S$, $i=1,2$, is surjective. Let the kernels of these two projections be $A, B\lhd R$, so that
    $A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$ and $B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$.
    Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely many coordinates.
  • $A$ and $B$ are Jacobson ideals of $R$. This is because $R/A\cong S$, $R/B\cong S$, and $S$ has trivial Jacobson radical.
  • Let $L''$ be the subring of $R$ consisting of all pairs $(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$. I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$. It is clear that $L''\cong L$.
  • Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
    I claim that $C$ is the kernel of a retraction of $R$ onto $L''$. To see this, first consider the intersection $C\cap L''$: since pairs of tuples in $C$ are zero almost everywhere and $L''$ is the diagonal copy of $L$ in $R$, the only common pair must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.

    Now I want to show that $R=C+L''$. Choose any $x=(\textbf{u},\textbf{v})\in R$ and define $\ell:=u_{\infty}=v_{\infty}$. The pair $d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$ while the difference $x-d\in R$ is zero almost everywhere, hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that $C$ is the kernel of a retraction of $R$ onto $L''\cong L$. (That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a transversal for $C$.)

added 7 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
added 1159 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
added 42 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
added 312 characters in body
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86
Source Link
Keith Kearnes
Keith Kearnes
  • 14.6k
  • 2
  • 50
  • 86