geometry ag.algebraica – In relation to local full intersection morphism

Following Qing Liu's book Algebraic geometry and arithmetic curvesIn section 6.3.2, we say that a map. $ f: X rightarrow Y $ of varieties on some field $ k $ is local full intersection in $ x in X $ if there is a neighborhood $ U $ of $ x $ and a chain of morphisms. $ U overset {i} { hookrightarrow} Z overset {g} { rightarrow} Y $, where $ g circ i = f | _U $, $ Z $ it's a scheme about $ k $, $ i $ it is a regular inlay and $ g $ It is a smooth map. We say $ f $ is local full intersection (l.c.i for short) if it is l.c.i in each $ x in X $.

Now, let me come to my problem. Leave $ X $ Being a smooth complex projective variety, $ Y $ be another complex projective variety (not necessarily soft), and leave $ f: X rightarrow Y $ Be a map. Leave $ Z subset Y $ Be a soft closed sub-variety. Leave $ f: X rightarrow Y $ Be a map with the property that. $ f $ it's an isomorphism about $ Y setminus Z $, plus $ f $ is a $ mathbb {P} ^ n $– package about $ Z $ for some $ n $.

I believe that $ f $ it is a morphism of l.c.i, for which I have a discussion, but I am not completely sure if it is correct. Can someone please stop by my discussion and let me know if it makes sense or not?

Claim: $ f $ It is a morphism of l.c.i.

Test: According to the Stacks project {}, the morphism of lci is fpqc-local at the base, and since any zariski cover is also a fpqc cover, enough for Show that $ f $ is l.c.i on a related deck of $ Y $.
We consider two cases:

1) yes $ y in Z $, choose a related neighborhood $ U $ of $ and $ such that $ f ^ {- 1} (U) cong U times mathbb {P} ^ n $ Y $ f | _ {f ^ {- 1} (U)} $ is the usual projection on the 1st component (remember that $ f $ It is a projective package $ Z $). As $ mathbb {P} ^ n $ has finished l.c.i. $ mathbb {C} $Y $ U $ it's flat on $ mathbb {C} $, we have that $ U times mathbb {P} ^ n rightarrow U $ it is also l.c.i due to the fact that a flat base change of a map l.c.i is a map l.c.i (see the Battery Project, From there we conclude that $ f ^ {- 1} (U) rightarrow U $ It is the map of l.c.i.

2) yes $ y in Y setminus Z =: V $, then cover $ V $ by affinity it opens, and since then $ f $ is isomorphism about $ V $, clearly $ f $ is a map of l.c.i on each of the open affinities.

From there we conclude that $ f $ It is the morphism of l.c.i. (shown)

Can someone please tell me if this is correct? Thanks in advance.

Theory of complexity: problem of simple interretation in relation to the polynomial hierarchy?

There is a logical interpretation of the different levels of the polynomial hierarchy, which extends the characterization of witnesses $ mathsf {NP} $ Y $ mathsf {coNP} $.

A language $ L $ is in $ Sigma_k ^ P $ if there is a polytime predicate $ f $ and a polynomial $ ell $ such that
x in L Leftrightarrow exists | y_1 | le ell (| x |) ; forall | y_2 | le ell (| x |) ; cdots Q | y_k | le ell (| x |) ; f (x, y_1, ldots, y_k).


  • $ exists | and | le ell (| x |) $ means that there is a number $ and $ whose length is at most $ ell (| x |) $ such that …
  • $ forall | and | le ell (| x |) $ means that for all $ and $ whose length is at most $ ell (| x |) $, the following holds …
  • $ Q $ is $ exists $ Yes $ k $ it is strange and $ for all $ Yes $ k $ even.

Similary, $ L $ is in $ Pi_k ^ P $ If you can write similarly, you only start with $ for all $.

As an example, $ mathsf {NP} ^ { mathsf {NP}} $ is $ Sigma_2 ^ P $, and it consists of all the languages ‚Äč‚Äčthat
x in L Leftrightarrow exists | y_1 | leq ell (| x |) ; forall | y_2 | leq ell (| x |) ; f (x, y_1, ldots, y_k).

As another example, $ mathsf {coNP} ^ { mathsf {NP}} $ is $ Pi_2 ^ P $.

Your third example is $ mathsf {P} ^ { mathsf {NP}} $, which $ Delta_2 ^ P $. I'm not sure what the logical characterization is.

General expression for the relation of polynomials of equal degree and nonzero coefficients

Leave $ {a_k } $ Y $ {b_k } $ Be sequences of real and positive constants and $ x in (0,1) $. Is there a known expression for the following relationship?
frac { sum_ {k = 0} ^ na_kx ^ k} { sum_ {k = 0} ^ nb_kx ^ k} = frac {a_n} {b_n} + frac {a_n} {b_n} left ( frac {a_ {n-1}} {a_n} – frac {b_ {n-1}} {b_n} right) x ^ {n-1} + cdots

sql – Convert column of type LONG in VARCHAR2 in ORACLE – problem related to the size of column value of type LONG in relation to VARCHAR2

(p_owner VARCHAR2, p_table VARCHAR2, p_column VARCHAR2, p_rowid UROWID) RETURN VARCHAR2 IS

v_cursor INTEGER; --cursor
v_length INTEGER; - Length of the length
v_tamanho INTEGER; - Size of what was returned.
v_sql VARCHAR (2000); - He passed
v_max VARCHAR2 (32760); - variable varchar2 with maximum length
v_clob CLOB; - clob type variable
v_long LONG; - variable of long type

v_cursor: = dbms_sql.open_cursor;

v_sql: = & # 39; SELECT & # 39; || p_column || & # 39; DE & # 39; || p_owner || & # 39; & # 39; || p_tabela || & # 39; WHERE ROWID =: row_id & # 39 ;;

OPEN v_syscur FOR v_sql USING p_rowid;
FETCH v_syscur INTO v_long;
v_length: = LENGTH (v_long);
CLOSE v_syscur;
dbms_sql.parse (v_cursor, v_sql, dbms_sql.NATIVE);
dbms_sql.bind_variable (v_cursor, & # 39; row_id & # 39;, p_rowid);
dbms_sql.define_column_long (v_cursor, 1);

IF (dbms_sql.execute_and_fetch (v_cursor) = 1) THEN
dbms_sql.column_value_long (v_cursor, 1, v_length, 0, v_max, v_tamanho);

dbms_lob.createtemporary (v_clob, FALSE, dbms_lob.CALL);
v_clob: = v_max;



Normally, with this function above Oracle, I convert a column of type LONG to VARCHAR2, however, there are cases in which the value of the LONG column exceeds the allowed range of VARCHAR2, which inevitably causes an error. How could I solve this problem? That is, to make the entire value of the LONG column be converted to VARCHAR2?

Algorithms – How do we guess the recurrence relation of the given equation?

In clrs, I've been reading about a method called substitution method to solve recurrence, here the author guessed that the solution is T (n) = O (n log n), but I'm confused, why not O (n ^ 2)? Is there a method to correctly guess the solution? Is there anything that I missed in this method?

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Different coordinates in mapbox maps in relation to Google maps?

I have changed a directory site with google maps to mapbox map, due to the google payment policy … Everything is fine, but the same poi coordinates in google maps are in different places in mapbox maps? Why? Is there any formula to solve this problem?

What are the changes made to the DNS servers of the gTLD registry in relation to my domain?

If you use name servers out of service (as in your test case) and do not use DNSSEC (did not say) then you are right, the 3 NS the records will be the only things inserted in the main zone related to your domain.

one) Servers of names outside Bailiwick

While using name servers that are not "in" their own zone, you will be using name servers that are not in bailiwick, and therefore, the registry does not need to publish glue records.

On the contrary, if you decide to use as a name server for example.comThen, to solve the egg and chicken problem that you just created, the registry should also publish a record of glue for which basically means one or several A or AAAA records for this name server. Of course, this must be done through the registrar.

And later you will need to make sure that this is updated each time the IP addresses of this server change, otherwise important problems will occur.


If you decide to activate DNSSEC on your domain, then the main zone must publish one or more DS records for your zone. This is done again through the registrar but it also depends on what the registry wants to receive (basically a DS or a DNSKEY, in the latter case it means that the registry generates the DS itself of the DNSKEY), and what algorithms allow (that change your list of options for your DS / DNSKEY records).

The registry will also publish RRSIG records in its DS registry (s), to allow the entire chain of trust that resolution validators can verify in each DNS exchange.

The NS records themselves are not covered by the RRSIG records, regardless of the fact that they are in the AUTHORITY section (but they are signed within the children's area, of course). And the glue records, that is. A+AAAA records corresponding to the objectives of the NS the records are in the ADDITIONAL section, and so unattached RRSIG records too.

normalization – Is a relation with 2 attributes always in BCNF?

I went through some tutorials about BCNF and I got a relationship with 2 attributes always in BCNF. My question is:
What happens if a relation with 2 attributes has no prime or non-prime attributes? That is, attributes A and B do not determine each other and are simply 1NF. From 2NF, the question of the main and non-main attributes appears. Any body there to clarify the matter. Thanks in advance.

How to express an empty relation (or table) in PostgreSQL?

Is there a way to express an empty relation in SQL without actually creating an empty table?

For testing purposes, I want to find a way to prove something like:

SELECT the count (*) FROM (...) my_empty_table;

where ... Generates an empty table online. (I hope to get 0 as a reply)

it's possible?

I tried without success:

SELECT the account (*) FROM () t;

, which give syntax errors.


=> SELECT count (*) FROM (SELECT) t;
(1 row)

(I'm mainly using SQL, but I'm also curious if there is a standard way to write this).