# Are there any non-trivial automorphisms of the field Q(sqrt2)?

• MathematicalPhysicist
In summary, the conversation discusses finding the number of automorphisms for the field Q(sqrt2) and proving that f:R->R has only the identity automorphism. The individual uses a polynomial and fixed point theorem to come to the conclusion that the only automorphism is the identity function. However, it is mentioned that R\{a} is not a field and transfinite induction should not be used. The conversation also brings up the concept of symmetries and how they can be used to find automorphisms.
MathematicalPhysicist
Gold Member
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 $$P(x)=b_0+b_1x+...+b_nx^n$$
and $$f(P(x))=f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n$$
they have the same degree, so when f(P(x))=0=P(x)
we have: $$f(b_0)+f(b_1)f(x)+...+f(b_n)f(x)^n=b_0+...+b_nx^n$$
from here iv'e concluded that it has only the identity function as an automorphism, am i right?

Last edited:
Why have you concluded that from there?

cause f(b_i)f(x)^i=b_ix^i
am i wrong here?

ok, iv'e looked in mathworld, i think i got my answer.

i have another question.
i need to prove that f:R->R has only the indentity automorphism.
well bacause f preserves order, by the fixed point theorem we can find a point a in R suhc that f(a)=a, now if reuse the automorphism on the field R\{a} we can reuse this theorem and thus by reiterating we can get that f is the identity function, am i right in this way?

loop quantum gravity said:
cause f(b_i)f(x)^i=b_ix^i
am i wrong here?

Yes. f is an isomorphism that maps Q to Q. It does not (necessarily) send x to x for all x, merely for x in Q.

loop quantum gravity said:
i have another question.
i need to prove that f:R->R has only the indentity automorphism.
well bacause f preserves order, by the fixed point theorem we can find a point a in R suhc that f(a)=a, now if reuse the automorphism on the field R\{a} we can reuse this theorem and thus by reiterating we can get that f is the identity function, am i right in this way?

R\{a} is not a field. R also does not contain a finite number (or even a countable number) of points, and I don't believe you're supposed to use transfinite induction...

so how to prove that its only automorphism is the identity function?

automorphisms are symmetries.

consider the case of the roots of x^2+1 = 0.

do you see any symmetries among the roots of this one? or of the root field C?

the symmetry you find should leave the real numbers fixed.

## 1. What are automorphisms of Q(sqrt2)?

Automorphisms of Q(sqrt2) are functions that preserve the algebraic structure of the field Q(sqrt2). In other words, they are bijective maps from Q(sqrt2) to itself that preserve addition, multiplication, and inverses.

## 2. How many automorphisms does Q(sqrt2) have?

Q(sqrt2) has exactly two automorphisms: the identity map (which leaves all elements unchanged) and the map that sends sqrt2 to -sqrt2 and vice versa.

## 3. What is the significance of automorphisms of Q(sqrt2)?

Automorphisms of Q(sqrt2) play an important role in the study of algebraic extensions and Galois theory. They also have applications in fields such as cryptography and coding theory.

## 4. Can automorphisms of Q(sqrt2) be extended to other number fields?

No, automorphisms of Q(sqrt2) cannot be extended to other number fields because they are specific to the field Q(sqrt2) and do not necessarily preserve the structure of other fields.

## 5. How are automorphisms of Q(sqrt2) related to the Galois group of sqrt2 over Q?

The automorphisms of Q(sqrt2) form a subgroup of the Galois group of sqrt2 over Q. In fact, the Galois group is isomorphic to the group of automorphisms. This relationship provides a powerful tool for studying the properties of Q(sqrt2) and its extensions.

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