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... Consider the system of equations f (x, y) = 0, g(x, y) = 0 where f (x, y) = 3x2 + 2xy + 3y 2 − 2 g(x, y) = 3x2 − 2xy + 3y 2 − 2 We will consider f and g as polynomials in x whose coefficients are functions of x. What this means can be seen by writing f and g as (3)x2 + (2y)x + (3y 2 − 2) (3)x2 + (−2 ...
... Consider the system of equations f (x, y) = 0, g(x, y) = 0 where f (x, y) = 3x2 + 2xy + 3y 2 − 2 g(x, y) = 3x2 − 2xy + 3y 2 − 2 We will consider f and g as polynomials in x whose coefficients are functions of x. What this means can be seen by writing f and g as (3)x2 + (2y)x + (3y 2 − 2) (3)x2 + (−2 ...
Solutions - Dartmouth Math Home
... Because R is an integral domain, we conclude that either rk − 1 = 0 or rm = 0. If rm = 0, then it follows (since R is an integral domain) that r = 0. But this is impossible as r was assumed to be nonzero. Therefore we must have rk − 1 = 0, that is, rk = 1. This implies r · rk−1 = 1. This shows that ...
... Because R is an integral domain, we conclude that either rk − 1 = 0 or rm = 0. If rm = 0, then it follows (since R is an integral domain) that r = 0. But this is impossible as r was assumed to be nonzero. Therefore we must have rk − 1 = 0, that is, rk = 1. This implies r · rk−1 = 1. This shows that ...
Example sheet 4
... Let R be a Noetherian ring. Show that the R-module Rn satisfies condition (N ), and hence that any finitely generated R-module satisfies condition (N ). 3. Let M be a module over an integral domain R. An element m ∈ M is a torsion element if rm = 0 for some non-zero r ∈ R. (i) Show that the set T of ...
... Let R be a Noetherian ring. Show that the R-module Rn satisfies condition (N ), and hence that any finitely generated R-module satisfies condition (N ). 3. Let M be a module over an integral domain R. An element m ∈ M is a torsion element if rm = 0 for some non-zero r ∈ R. (i) Show that the set T of ...
Lecture Notes 13
... It is clear that this algorithm halts exactly on all inputs in L which means that L is semidecidable. Now we notice that L is not just semidecidable but also a decidable set. Indeed, consider another algorithm. It attempts to divide the integer number c by the integer number a using, for example, th ...
... It is clear that this algorithm halts exactly on all inputs in L which means that L is semidecidable. Now we notice that L is not just semidecidable but also a decidable set. Indeed, consider another algorithm. It attempts to divide the integer number c by the integer number a using, for example, th ...
Unit One Combined Notes
... An integer is all __________ and __________ numbers, excluding ___. Numbers such as (+16) and (-12) are _____________. (+16) is a ___________ integer (-12) is a ___________ integer We can use tiles to represent integers ...
... An integer is all __________ and __________ numbers, excluding ___. Numbers such as (+16) and (-12) are _____________. (+16) is a ___________ integer (-12) is a ___________ integer We can use tiles to represent integers ...
June 2007 901-902
... 3. Find, with justification, the complete character table for S4 , the permutation group on 4 letters. (There are many ways of doing this, but here is one tip that might help: Let V = Ce1 ⊕ Ce2 ⊕ Ce3 ⊕ Ce4 be a four-dimensional vector space over C. Consider V as a C[S4 ]-module by defining σei := eσ ...
... 3. Find, with justification, the complete character table for S4 , the permutation group on 4 letters. (There are many ways of doing this, but here is one tip that might help: Let V = Ce1 ⊕ Ce2 ⊕ Ce3 ⊕ Ce4 be a four-dimensional vector space over C. Consider V as a C[S4 ]-module by defining σei := eσ ...
