Hello Experts,
I am given that d = 1 mod 4.
I need to show that Z[sqrt(d)] isn't Unique Factorization Domain.
Also I need you to explain me why 1+sqrt(d) is irreducible.
What I did:
Take a = 1+sqrt(d)
a^2 = 1 + d + 2*sqrt(d)
I know that 2 is irreducible in Z[sqrt(d)].
a^2 = a*a = (1+sqrt(d))*(1+sqrt(d)) = 2*(n+sqrt(d))
So I need to determine why 1+sqrt(d), n+sqrt(d) are irreducible and why Z[sqrt(d)] isn't UFD.
I am given that d = 1 mod 4.
I need to show that Z[sqrt(d)] isn't Unique Factorization Domain.
Also I need you to explain me why 1+sqrt(d) is irreducible.
What I did:
Take a = 1+sqrt(d)
a^2 = 1 + d + 2*sqrt(d)
I know that 2 is irreducible in Z[sqrt(d)].
a^2 = a*a = (1+sqrt(d))*(1+sqrt(d)) = 2*(n+sqrt(d))
So I need to determine why 1+sqrt(d), n+sqrt(d) are irreducible and why Z[sqrt(d)] isn't UFD.
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When trying to prove something, it is often helpful to look for counter-examples. That search may help you understand why they don't exist, leading to a proof. Or you might find an counter-example, showing that no proof is possible!
In this case, consider d=17 and multiply (4+√17)(13-3√17).
Fortunately, you do not need 1+√d to be irreducible to show that Z[√d] isn't a UFD. Since you know that "2" is irreducible, it is enough to show that 2 does not factor 1+√d. For integers "n" and "m", can 2(n+m√d)=1+√d ? The expression (1+√d)*(1+√d) might not be completely factored, but if you continue to factor it into irreducible factors {that is factor (1+√d) into irreducible factors and list them all twice} can "2" ever be among those factors? So the factoring you have presented always gives (possibly with some further factoring) two different irreducible factorings (2 is included on one side, but cannot be included on the other). Irreducible Factorizations are not unique: Z[√d] isn't a UFD.
In this case, consider d=17 and multiply (4+√17)(13-3√17).
Fortunately, you do not need 1+√d to be irreducible to show that Z[√d] isn't a UFD. Since you know that "2" is irreducible, it is enough to show that 2 does not factor 1+√d. For integers "n" and "m", can 2(n+m√d)=1+√d ? The expression (1+√d)*(1+√d) might not be completely factored, but if you continue to factor it into irreducible factors {that is factor (1+√d) into irreducible factors and list them all twice} can "2" ever be among those factors? So the factoring you have presented always gives (possibly with some further factoring) two different irreducible factorings (2 is included on one side, but cannot be included on the other). Irreducible Factorizations are not unique: Z[√d] isn't a UFD.