2020-2ν•™κΈ°, λŒ€ν•™μ—μ„œ β€˜ν˜„λŒ€λŒ€μˆ˜1’ μˆ˜μ—…μ„ λ“£κ³  κ³΅λΆ€ν•œ λ°”λ₯Ό μ •λ¦¬ν•œ κΈ€μž…λ‹ˆλ‹€. 지적은 μ–Έμ œλ‚˜ ν™˜μ˜μž…λ‹ˆλ‹€ :)

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2020-2ν•™κΈ°, λŒ€ν•™μ—μ„œ β€˜ν˜„λŒ€λŒ€μˆ˜1’ μˆ˜μ—…μ„ λ“£κ³  κ³΅λΆ€ν•œ λ°”λ₯Ό μ •λ¦¬ν•œ κΈ€μž…λ‹ˆλ‹€. 지적은 μ–Έμ œλ‚˜ ν™˜μ˜μž…λ‹ˆλ‹€ :)


Artin, p.442, sec.2, #2a.

The ring Z[ΞΆ], for ΞΆ=e2Ο€i/3, is a Euclidean Domain.


The Norm function N would be complex modulus |β‹…| square.

  1. Check N is integer valued on Z[ΞΆ].
  2. The Division Algorithm works.


(1) N is integer valued.

ΞΆλŠ” root of unity 쀑 ν•˜λ‚˜ μ΄λ―€λ‘œ μ•„λž˜κ°€ μ„±λ¦½ν•œλ‹€.

  • ΞΆ3=1
  • ΞΆ2=1βˆ’ΞΆ

λ”°λΌμ„œ Z[ΞΆ]의 μ›μ†ŒλŠ” μ•„λž˜μ™€ 같은 ν˜•νƒœλ‘œ κΈ°μˆ λœλ‹€.

a+b΢for somea,b∈Z

(1이 βŸ¨ΞΆβŸ©μ— ν¬ν•¨λ˜μ–΄ 있기 λ•Œλ¬Έ!)

λ˜ν•œ, ΞΆ=βˆ’12+32iμ΄λ―€λ‘œ, Norm N은

|a+bΞΆ|2=|(aβˆ’b2)+3b2i|2=a2βˆ’ab+b24+3b24=a2βˆ’ab+b2β‰₯(aβˆ’b)2β‰₯0

λ”°λΌμ„œ N은 integer valued function이닀!


(2) Division Algorithm

For a,b∈Z[ΞΆ] and aβ‰ 0, check that there exist q,r∈Z[ΞΆ] s.t.

b=qa+r

where N(r)<N(a).

Let q be the closest point to b/a on Z[ΞΆ].

Let r be r=bβˆ’qa.

Then check that

|r|=|bβˆ’qa|=|baβˆ’q||a|

μ΄λ•Œ, |baβˆ’q|에 λŒ€ν•΄ μƒκ°ν•΄λ³΄μž.

baλ₯Ό μ œλŒ€λ‘œ κ΅¬ν•˜λ©΄, ba=s+t΢∈Q[ΞΆ]이닀.

μ΄λ•Œ qλ₯Ό ba에 κ°€μž₯ κ°€κΉŒμš΄ Z[ΞΆ]둜 μ„€μ •ν–ˆμœΌλ―€λ‘œ

q=x+yΞΆ,where|xβˆ’s|≀12and|yβˆ’t|≀12

λ”°λΌμ„œ |baβˆ’q|λŠ”

|baβˆ’q|2=|(s+tΞΆ)βˆ’(x+yΞΆ)|2=|(sβˆ’x)+(tβˆ’y)ΞΆ|2=|((sβˆ’x)βˆ’(tβˆ’y)2)+(tβˆ’y)32i|2=(sβˆ’x)2βˆ’(sβˆ’x)(tβˆ’y)+(tβˆ’y)24+3(tβˆ’y)24=(sβˆ’x)2βˆ’(sβˆ’x)(tβˆ’y)+(tβˆ’y)2≀14+14+14=34

λ”°λΌμ„œ

|r|=|bβˆ’qa|=|baβˆ’q||a|≀34|a|<|a|

즉, Division Algorithm이 μ„±λ¦½ν•˜λ―€λ‘œ Z[ΞΆ]λŠ” Euclidean Algorithm이닀. β—Ό



μ•žμ„  λ¬Έμ œλ„ Z[βˆ’2]κ°€ Euclidean Domainμž„μ„ λ³΄μ΄λŠ” λ¬Έμ œμ˜€λŠ”λ°, μ΄λ²ˆμ—λ„ λΉ„μŠ·ν•œ λ¬Έμ œμ˜€λ‹€.

문제λ₯Ό ν’€κ³  λ‚˜μ„œ λ“  생각은 μ§€κΈˆμ˜ νŒ¨ν„΄μ„ μΌλ°˜ν™”ν•˜λ©΄ Z[Ο‰] for any Ο‰βˆˆCλ₯Ό Euclidean Domain으둜 λ§Œλ“€ 수 μžˆμ„μ§€ 생각해봀닀.

그런데 Euclidean Domain에 λŒ€ν•΄ 검색을 μ’€ ν•΄λ³΄λ‹ˆ μ•„λž˜μ™€ 같은 쑰건 μ•„λž˜μ—μ„œ Z[Ο‰]κ°€ Euclidean Domain이 λ˜λŠ” 것 κ°™λ‹€.


Case 1: Z[βˆ’n]

If n=1 or n=2, then Z[βˆ’n] is an Euclidean Domain.

If nβ‰₯3, then Z[βˆ’n] is not an Euclidean Domain.

일단 이건 baβˆ’q에 λŒ€ν•œ Norm을 ꡬ해보면, 1+n4κ°€ λ‚˜μ˜€λŠ”λ°, 이게 1+n4<1이 λ˜μ•Ό ν•΄μ„œ 그런 걸둜 μ•Œκ³  μžˆλ‹€.


Case 2: Z[Ο‰]

이 κ²½μš°μ—λŠ” Ο‰κ°€ cube root of unityκ°€ λ λ•Œ Euclidean Domain이 λœλ‹€.

Ο‰κ°€ cube root of unityκ°€ λ˜λŠ” κ²½μš°λŠ” Ο‰=e2Ο€i/3κ³Ό Ο‰=e4Ο€i/3이닀.


κ·Έ μ™Έμ˜ κ²½μš°λŠ” λ­”κ°€ 직접 baβˆ’qλ₯Ό ꡬ해봐야 ν•  것 κ°™λ‹€. μ•„λ‹˜ λ‚΄κ°€ 아직 μ°Ύμ§€ λͺ»ν•œ Ο‰βˆˆC에 λŒ€ν•œ 쑰건이 μžˆμ„ μˆ˜λ„?