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MathematicalPhysicist

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i need to find how many automorphisms are there for the field Q(sqrt2).

i am given the hint to use the fact that a is a root of a polynomial on Q(sqrt2).

what i did is:

if f:Q(sqrt2)->Q(sqrt2) is an automorphism then if P(x) is a polynomial on this field then:

if a is a root of P(x), then P(a)=0 but then also f(P(a))=f(0)=0

so [tex]P(x)=b_0+b_1x+...+b_nx^n[/tex]

and [tex]f(P(x))=f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n[/tex]

they have the same degree, so when f(P(x))=0=P(x)

we have: [tex]f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n=b_0+...+b_nx^n[/tex]

from here iv'e concluded that it has only the identity function as an automorphism, am i right?

i am given the hint to use the fact that a is a root of a polynomial on Q(sqrt2).

what i did is:

if f:Q(sqrt2)->Q(sqrt2) is an automorphism then if P(x) is a polynomial on this field then:

if a is a root of P(x), then P(a)=0 but then also f(P(a))=f(0)=0

so [tex]P(x)=b_0+b_1x+...+b_nx^n[/tex]

and [tex]f(P(x))=f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n[/tex]

they have the same degree, so when f(P(x))=0=P(x)

we have: [tex]f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n=b_0+...+b_nx^n[/tex]

from here iv'e concluded that it has only the identity function as an automorphism, am i right?

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