The lattice of ideals of a distribituve lattice is itself distributive The 2019 Stack Overflow Developer Survey Results Are InThere's in this relation a distributive lattice?a question about distributive latticeA question related to distributive bounded lattice.Elements of bounded distributive lattice belonging to same prime ideals are equal?“One cannot hope to find any further essentially new lattice properties…”Is it true that a distributive lattice with finite length is finitelattice theory questions about idealsDistributive latticesThe lattice of annihilator ideals of a ringWhy wouldn't the distributive laws hold for a lattice?
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The lattice of ideals of a distribituve lattice is itself distributive
The 2019 Stack Overflow Developer Survey Results Are InThere's in this relation a distributive lattice?a question about distributive latticeA question related to distributive bounded lattice.Elements of bounded distributive lattice belonging to same prime ideals are equal?“One cannot hope to find any further essentially new lattice properties…”Is it true that a distributive lattice with finite length is finitelattice theory questions about idealsDistributive latticesThe lattice of annihilator ideals of a ringWhy wouldn't the distributive laws hold for a lattice?
$begingroup$
I have found myself stuck on a problem and would appreciate a hint. The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive. This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.
lattice-orders
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
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add a comment |
$begingroup$
I have found myself stuck on a problem and would appreciate a hint. The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive. This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.
lattice-orders
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
$endgroup$
add a comment |
$begingroup$
I have found myself stuck on a problem and would appreciate a hint. The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive. This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.
lattice-orders
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
$endgroup$
I have found myself stuck on a problem and would appreciate a hint. The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive. This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.
lattice-orders
lattice-orders
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
asked Apr 8 at 0:41
Jean-Pierre de VilliersJean-Pierre de Villiers
536
536
add a comment |
add a comment |
1 Answer
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Hint: to show that
$$(I vee J) wedge K = (I wedge K) vee (J wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I vee J = a in L : exists iin I;exists j in J;(a leq i vee j) .$$
Additionally, you might find useful the fact that if $k leq i vee j$ then $k = k wedge (i vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
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1 Answer
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1 Answer
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$begingroup$
Hint: to show that
$$(I vee J) wedge K = (I wedge K) vee (J wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I vee J = a in L : exists iin I;exists j in J;(a leq i vee j) .$$
Additionally, you might find useful the fact that if $k leq i vee j$ then $k = k wedge (i vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
$endgroup$
add a comment |
$begingroup$
Hint: to show that
$$(I vee J) wedge K = (I wedge K) vee (J wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I vee J = a in L : exists iin I;exists j in J;(a leq i vee j) .$$
Additionally, you might find useful the fact that if $k leq i vee j$ then $k = k wedge (i vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
$endgroup$
add a comment |
$begingroup$
Hint: to show that
$$(I vee J) wedge K = (I wedge K) vee (J wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I vee J = a in L : exists iin I;exists j in J;(a leq i vee j) .$$
Additionally, you might find useful the fact that if $k leq i vee j$ then $k = k wedge (i vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
$endgroup$
Hint: to show that
$$(I vee J) wedge K = (I wedge K) vee (J wedge K),$$
where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and
$$I vee J = a in L : exists iin I;exists j in J;(a leq i vee j) .$$
Additionally, you might find useful the fact that if $k leq i vee j$ then $k = k wedge (i vee j)$, and use distributivity.
I don't think that I can give you any more hints without spoiling it. Good luck!
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
answered Apr 8 at 8:45
amrsaamrsa
3,8702618
3,8702618
add a comment |
add a comment |
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