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# Euclidean Domains watch

1. How does one show if Z[rt2] is/is not (think is) a euclidean domain?
2. It's a euclidean domain because it has euclidean valuation/gauge function defined by .
3. (Original post by fabuleuse)
It's a euclidean domain because it has euclidean valuation/gauge function defined by .
No as -1 + rt(2) is a unit so V(-1 + rt(2)) has to be the same as V(1)
I believe the function to be V(a+b rt(2)) = a^2 - 2b^2 but showing the divsion algo works with this function is tricky
4. Let x = rt(2) for ease of notation.

Check d(a+bx) = |a^2 - 2b^2| is multiplicative on Z[x]. Also I will extend this d to Q[x].

Now if p and q are rationals then choosing integers a and b near (within 1/2) of p and q gives

d((p+qx) - (a+bx)) = |(p-a)^2 - 2(q-b)^2| =< 2(1/2)^2 = 1/2 < 1.

So if H and K are two elements of Z[x], and noting H/K is an element of Q[x], then we can find L in Z[x] such that

d(H/K - L) < 1.

As d is multiplicative then

d(H - KL) < d(K)

as required of the division algorithm.
5. gota love the mathmos

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