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Gaussian and Eisenstein Integers (Part 1)

In this video we have a look at Gaussian (G) and Eisentein (E) integers. With respect to complex number addition and multiplication, G and E are each a subring of the field of complex numbers and E and G are therefore each an integral domain. In this video we prove that G and E are each a Euclidean domain if we take the usual Norm of a complex number as the Euclidean function. We do this by contradiction by showing that for each ring there is at least one element h such that N(f-gh) is less than N(f) whenever f and g are elements of one ring and N(f) is greater than or equal to N(g) which is nonzero. For the Gaussian integers we show that h is at least one of 1,-1, i, -i. Notice that these are the four fourth roots of unity and that these are actually all the units (elements with a multiplicative inverse) in the ring of Gaussian integers. Similarly for the Eisenstein integers we show that h is at least one of 1,-1,ω,-ω,1+ω,-1-ω which are the six sixth roots of unity and all the units in the ring of Eisenstein integers. Timestamps 00:00 Definitions 01:21 cube root of 1 03:40 Norm 08:04 Gaussian Integers 14:31 Eisenstein Integers At 01:21, for example if ω=-0.5+i(\sqrt{2}/2)i, then we can change a to a-b and b to -b and then we get that a+bω becomes a-b-bω=a+bω² because 1+ω+ω²=0. At 06:50 we have made use of the following result: Let R be an integral domain and let N be a function defined on the set of nonzero elements of R and having values in the set of nonnegative integers. Suppose the following condition is satisfied: For each pair of nonzero elements f,g in f for which N(f) \geq N(g)
eq 0, then there exists an element h of R such that either f=gh or N(f-gh)\geq N(f). Then N is a Euclidean function and R is a Euclidean domain. At 11:27 and 19:38 the 'implies' sign was meant rather than the 'is equivalent to' sign because of course it is the implication that is true.