Teorema prostih brojeva — разлика између измена

== Analog za nesvodljive polinome na konačnom polju ==
 
Postoji analogna teorema prostih brojeva koja opisuje „raspodelu” [[irreducible polynomial|nesažimljivih polinoma]] preko [[Коначно поље|konačnog polja]]; njen oblik je upadljivo sličan sa klasičnom teoremom prostih brojeva.
There is an analogue of the prime number theorem that describes the "distribution" of [[irreducible polynomial]]s over a [[finite field]]; the form it takes is strikingly similar to the case of the classical prime number theorem.
 
ToDa statebi itse preciselyto precizno izrazilo, letmože se uzeti da je {{math|''F'' {{=}} GF(''q'')}} bekonačno thepolje finite field withsa {{mvar|q}} elementselemenata, forza someneko fixedfiksno {{mvar|q}}, andi letda je {{mvar|N<sub>n</sub>}} be the number ofbroj [[monic polynomial|monicmonijskih]] ''irreducible''nesažimljivih polynomialspolinoma overpreko {{mvar|F}} whosečiji je [[degree of a polynomial|degreestepen]] is equal tojednak {{mvar|n}}. ThatTo isjest, werazmatraju arese lookingpolinomi atsa polynomialskoeficijentima withodabranim coefficients chosen fromiz {{mvar|F}}, whichkoji cannotse bene writtenmogu aszapisati productskao ofproizvodi polynomialspolinoma ofnižeg smaller degreestepena. InU thisovom settingokruženju, theseti polynomialspolinomi playigraju theulogu roleprostih of the prime numbersbrojeva, sincejer allsu othersvi monicdrugi polynomialsmonijski arepolinomi builtizgrađeni upod ofnjihovih productsproizvoda. of them.Onda Onese canmože thendokazati proveda thatje
:<math>N_n \sim \frac{q^n}{n}.</math>
IfAko wese makeuradi the substitutionsupstitucija {{math|''x'' {{=}} ''q''<sup>''n''</sup>}}, then the rightonda handje sidedesna isstrana justsamo
:<math>\frac{x}{\log_q x},</math>
whichčime makesse thepojašnjava analogy cleareranalogija. Since thereKako arepostoji preciselytačno {{math|''q''<sup>''n''</sup>}} monicmonijskih polynomialspolinoma of degreestepena {{mvar|n}} (includinguključujući theone reduciblekoji onessu sažimljivi), thisto canse bemože rephrasedpreformulirati asna followssledeći način: ifako aje monicmonijski polynomialpolinom of degreestepena {{mvar|n}} israndomno selected randomlyizabran, then theonda probabilityje ofverovatnoća itda beingje irreducibleon isnesažimljiv aboutoko&nbsp;{{math|{{sfrac|1|''n''}}}}.
 
Moguće je dokazati i analognu verziju Rimanove hipoteze, naime da je
One can even prove an analogue of the Riemann hypothesis, namely that
:<math>N_n = \frac{q^n}n + O\left(\frac{q^\frac{n}{2}}{n}\right).</math>
 
TheDokazi proofsovih oftvrdnji thesedaleko statementssu arejednostavniji farnego simpleru thanklasičnom in the classical caseslučaju. It involvesTo aobuhvata shortkratako combinatorialkombinatorično argumentrazmatranje,<ref>{{cite journal|last=Chebolu|first=Sunil|first2=Ján|last2=Mináč|title=Counting Irreducible Polynomials over Finite Fields Using the Inclusion {{pi}} Exclusion Principle|journal=Mathematics Magazine|date=December 2011|volume=84|issue=5|pages=369–371|doi=10.4169/math.mag.84.5.369|jstor=10.4169/math.mag.84.5.369|arxiv=1001.0409}}</ref> summarisedsumirano asna followssledeći način: everysvaki element of the degreestepena {{mvar|n}} extension ofproširenja {{mvar|F}} isje akoren rootnekog ofnesažimljivog somepolinoma irreduciblečiji polynomial whose degreestepen {{mvar|d}} dividesdeli {{mvar|n}}; by counting thesepri rootsprebrojavanu inovih twokorena differentsu waysuspostavljena onedva establishesrazličita thatpristupa
:<math>q^n = \sum_{d\mid n} d N_d,</math>
wheregde thesuma sumobuhvata is over allsve [[divisorДељивост|dilioce]]s {{mvar|d}} ofod {{mvar|n}}. [[Möbius inversion formula|Mebijusova inverzija]] thenonda yieldsdaje
:<math>N_n = \frac{1}{n} \sum_{d\mid n} \mu\left(\frac{n}{d}\right) q^d,</math>
wheregde je {{math|''μ''(''k'')}} is the [[MöbiusMebijusova functionfunkcija]]. (ThisOva formula wasje knownbila topoznata GaussGausu.<!-- although I haven't got a reference for this. -->) TheGlavni mainčlan termse occursjavlja forza {{math|''d'' {{=}} ''n''}}, andi itnije isteško notvezati difficultpreostale to bound the remaining termsčlanove. TheIzraz "Riemann„Rimanove hypothesis"hipoteze” statementzavisi dependsod ončinjenice theda fact that the largestnajveći [[properДељивост|svojstveni divisordililac]] ofod {{mvar|n}} canne može beda nobude largerveći thanod {{math|{{sfrac|''n''|2}}}}.
 
== Reference ==