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University of Paderborn Department of Mathematics Diploma Thesis ...

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44 CHAPTER 2. STABLE IDEALSBinomial representation: Polynomial representation: Sum <strong>of</strong> former polynomials:( ) z + 2p 1 (z) :=p 1 (z) = 1 22 z2 + 3 1∑2 z + 1 p i (z) = 1 2 z2 + 3 2 z + 1( ) z + 2 − 1p 2 (z) :=2( ) z + 2 − 2p 3 (z) :=2( ) z + 2 − 3p 4 (z) :=2( ) z + 1 − 4p 5 (z) :=1( ) z + 1 − 5p 6 (z) :=1( ) z + 0 − 6p 7 (z) :=0( ) z + 0 − 7p 8 (z) :=0( ) z + 0 − 8p 9 (z) :=0( ) z + 0 − 9p 10 (z) :=0( )z + 0 − 10p 11 (z) :=0( )z + 0 − 11p 12 (z) :=0p 2 (z) = 1 2 z2 + 1 2 zp 3 (z) = 1 2 z2 − 1 2 zp 4 (z) = 1 2 z2 − 3 2 z + 1p 5 (z) = z − 3p 6 (z) = z − 4p 7 (z) = 1p 8 (z) = 1p 9 (z) = 1p 10 (z) = 1p 11 (z) = 1p 12 (z) = 1i=12∑p i (z) = z 2 + 2z + 1i=13∑p i (z) = 3 2 z2 + 3 2 z + 1i=14∑p i (z) = 2z 2 + 2i=15∑p i (z) = 2z 2 + z − 1i=16∑p i (z) = 2z 2 + 2z − 5i=17∑p i (z) = 2z 2 + 2z − 4i=18∑p i (z) = 2z 2 + 2z − 3i=19∑p i (z) = 2z 2 + 2z − 2i=1∑10i=1∑11i=1∑12i=1p i (z) = 2z 2 + 2z − 1p i (z) = 2z 2 + 2zp i (z) = 2z 2 + 2z + 1This provides a 1 = a 2 = a 3 = a 4 = 2, a 5 = a 6 = 1 and a 7 = a 8 = a 9 = a 10 = a 11 = a 12 = 0in the terminology <strong>of</strong> Theorem 2.32. Hence, the upper bound for the degrees is 12. Indeed,the ideal L p := (x 0 , x 5 1, x 4 1x 3 2, x 4 1x 2 2x 6 3) ⊂ R contains one monomial generator <strong>of</strong> degree 12and we obtain p R/Lp (z) = 2z 2 + 2z + 1.

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