Find all possible ring homomorphisms on q\sqrt 2 to q\sqrt2
Answers
Answer:
Hint: If ϕ is a homomorphism then ϕ(2–√)2=ϕ(2). If ϕ is non-trivial then ϕ(2)=2...
Hint: If ϕ is a homomorphism then ϕ(2–√)2=ϕ(2). If ϕ is non-trivial then ϕ(2)=2...Oops. It's been pointed we need to show that ϕ(1)=1 if ϕ is nontrivial. In any case, if x=ϕ(1) then x2=x; since there are no zero divisors in Q[3–√] this shows that ϕ(1) is 0 or 1, and if ϕ(1)=0 then ϕ is trivial......
☺❤#⃣☺❤#⃣☺❤#⃣
In other words, if f is a ring homomorphism from Q[2–√] to Q[3–√], then f(r)=0 for all r∈Q[2–√]...
I was trying to see if the assumed f contradicts any one of axioms of ring homomorphisms, but didn’t see one. I was thinking using the fact that Q[2–√] and Q[3–√] are fields so the only ideals are trivial ideals. Since ker(Q[2–√]) is also an ideal, I want to show that this ker(Q[2–√]) is not the zero ideal. Then I didn’t go too far because I have no other information about the structure of these two fields. Also, I was thinking about representing any non-zero r in Q[2–√] and to map it and somehow hopefully use the no zero-divisor property of Integral Domain to show that r is mapped to 0…But I’m not sure if the specific Q[2–√] and Q[3–√] can give us stronger hypothesis…