Tag: campi
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The Diophantine equation x²-y³=7.
We will prove there are not integer solution to this equation. Suppose that (x,y) is a possible solution of the Diophantine equation. If y is even, reducing modulo 8 we would have x²=7 mod 8, which is absurd because 7 does not fit among the squares modulo 8. It follows that y is odd. Then…
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If m is odd and n a natural number (2ᵐ-1,2ⁿ+1)=1.
In the following i denote with (a,b) the greatest common divisor between two integers a and b. Let p now be a possible prime divisor of A=2ᵐ-1 and B=2ⁿ+1. Let [2]_p denote the class of 2 modulo p. Let us note that since (2,p)=1 then [2]_p is an element of the group of invertibles of…
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There is no group G such that G’=D2n, for n≥3.
Let G be a group. We denote with G’ the commutator subgroup of G and with D2n the diedral group of order 2n, with n a natural number ≥3. Let us prove that there is no group G such that G’=D2n. Suppose that such a group exists, then since G’=<r,s> we get: <r> char G’…
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If R is a local ring then R[[x]] is local.
Recall that if S is a commutative ring with unit, S is local if and only the set of non unit of S is an ideal, i.e, denoting with S* the unit of S, S-S* is an ideal of S. Now, let m be the maximal ideal of R. We prove that R[[x]] – R[[x]]*…
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Let R be a local ring and a,b in R such that (a)=(b). Prove that there exists a unit r of R such that a=rb.
We have a=rb and b=sa for some element r and s of R. Then a=rsa so a(1-rs)=0. Clearly if a=0 then b=0 and we can just take r=1. Suppose a≠0. If r∈R \ R* then, since R is local, r∈J(R) where J is the Jacobson radical of R. Then 1-rs ∈ R* so that a=0,…
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If f∈ℤ[x,y] generete only prime, then f is constant.
Suppose f is not constant, then, since ℤ[x,y]=(ℤ[y])[x] we can write: f=a+f₁(y)x + … + fk(y)xᵏ where fi ∈ ℤ[y] and fi≠0 for every i=1,…,k. I claim that there is y0∈ℤ such that f(x,y0) is not constant. Indeed if for every y0 in ℤ f(x,y0) is constant, then for every i=1,…,k fi(y) is zero in…
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Let F be a field. Can F be realized as a quotient of ℤ[i] ?
Lemma 1: If F ≅ ℤ[i]/J for some ideal J of ℤ[i] then F is a finite field. Proof: It is sufficient to show that ℤ[i]/J is finite. Since ℤ[i] is not a field we can assume J not equal to zero. Since ℤ[i] is a Euclidean domain it is in particular a PID so…
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Let K be an algebraically closed field and a,b n×n matrices over K. Then ab and ba have the same characteristics polynomial.
There are a lot of proof of this fact however here we want to gave a proof that use the machinery of Algebraic Geometry. Recall that we define the affine n space over K as Aⁿ=Kⁿ . This set can be endowed with a nice topology, namely the Zariski topology. This is the topology which…
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Let G be a finite group such that Aut(G)={e}. Prove that G is isomorphic to ℤ₂
Since Aut(G)={e} we would have that G is abelian (because we would have G/Z(G) ={e} and so G=Z(G) ) then the inversion x|—>x⁻¹ is an automorphism of G. If G has an element of order greater then 2 then the inversion is not trivial so G is a 2-elementary abelian groups and so G is…
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Solve in ℤ₅ the equation x⁷ + x³ – x +1= 0.
If x ∈ ℤ₅ such that x⁷+x³-x+1=0 then x(x⁶+x²-1)=-1 i.e. x(2x²-1)=4. Now (when I write x=m i mean the class of m modulo 5) x=0 —> x(2x²-1)=0≠4x=1 —> x(2x²-1)=1≠4x=2 —> x(2x²-1)=14=4x=3 —> x(2x²-1)=1≠4x=4 —> x(2x²-1)=124=4 so the solutions are the class of 2 and the class of 4 modulo 5.
Marco Damele 