I was wondering whether you can help me with some issues I am having:

Notation:

Aut(R)

is the set of automorphism of a ring R.

1) how can you formally show that Aut(R) is a “group under composition”? I am not quite sure what the question is asking to show!

2) the other day you gave me an easy way of showing ring isomorphisms, and was wondering how to show

Aut(Q[x]/((x^2)-p))

is isomorphic to C2 (cyclic group of order 2) and describe the nontrivial element explicitly (where p is prime)

3) not sure how to approach questions like this

Aut(F2[x]/((x^3)+x+1)) = ?

and

Aut(Q) is trivial

would i have to use Eisensteins criterion?

4) how can i show that any automorphism of reals is continuous and that

Aut(R)

is trivial?

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Also, would you be able to tell me how to compute the complete factorization into Q-irreducible factors for these polynomials:

i)a P[x] in the form of

(x^a + 1)

where a is a positive integer, eg (x^10 + 1) or (x^16 + 1)

ii) (x^5 + 10x^4 + 13x^3 – 25x^2 -68x – 60)

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Show that R[x]/(x^2 + a) is isomorphic to Complex if a>0 and RxR if a0 and RxR if a<0 where R is Real.

Reply:

(1) You’ve probably learned that the set of bijections of any set is a group under composition. Now when the set is a ring R, then Aut(R) is the subset of the group of bijections from R to R consisting of the maps that preserve the two laws of composition and the unit(s). Now what does it take to show that a subset is in fact a subgroup?

(2) To get a ring homomorphism from Q[x]/(x^2-p) to itself, you need to have a map

f:Q[x]->Q[x}

that takes x^2-p to x^2-p. Such a ring homomorphism is entirely determined by the value f(x). (Why?) But to preserve the polynomial, we must have f(x)=x or -x. Both of these will induce automorphisms of Q[x]/(x^2-p) .

Incidentally, depending on what you’ve learned and what’s expected in class, you might have to provide much more detail than I’ve written here. (The same goes for the other problems.)

(3) Eisenstein’s criterion, dealing with irreducibility of integer coefficient polynomials, is irrelevant here. But for polynomials of degree 2 or 3, a straightforward criterion for irreducibility is the non-existence of a root in the field. Use this to check that x^3+x+1 is irreducible. As a result, F_2[x]/(x^3+x+1) is a field extension of F_2 of degree 3. So the group of automorphisms is of order at most 3. Now, for a field of characteristic 2, the squaring map

a-> a^2

is a ring homomorphism. Think about this carefully. It might be convenient to note also that an endomorphism of a finite field is automatically bijective. (Why?)

To see the triviality of Aut(Q), note that any ring homomorphism f:Q->Q must have f(0)=0 and f(1)=1. But then, by additivity of the map, f(n)=n for any natural number. If you have a positive rational number r=n/m with m>0, then

f(mr)=mf(r)

on the one hand, and

f(mr)=f(n)=n

on the other. So

mf(r)=n

and

f(r)=n/m=r.

Now deal with the negative numbers in some obvious way.

(4) This is quite an interesting fact. The point is that a field automorphism of R must preserve the *order* of R. This is because

a>b

iff

a-b>0

iff

x^2-(a-b)

has a root in R. This is a property that is preserved by any field automorphism. Now can you check that an order-preserving map of R is necessarily continuous? Here, you might have to review the precise definition of continuity from basic analysis.

Once you have the continuity, the triviality of Aut(R) follows from the corresponding fact for Q and the *densenesss* of Q in R.

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(i) There is a recursive algorithm. A basic statement is that

x^p-1=(x-1)(x^{p-1}+x^{p-2}+…+x+1)

and

(x^{p-1}+x^{p-2}+…+x+1)

is irreducible. You can show this using the substitution x->x+1 and Eisenstein’s criterion. For other cases, do some research on cyclotomic polynomials, as you seem to have done already. Remember to look for the *recursive* algorithm, which is the most efficient way. These polynomials are very important in algebraic number theory, but also its applications to information theory.

(ii) You seem to have figured out already the substitution that reduces this to Eisenstein’s criterion. Unfortunately, I don;’t know any clever method for finding such substitutions. There might be other proofs of irreducibility for this polynomial, depending on what else you’ve learned.

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I hope you’ve learned already that C is isomorphic to R[x]/(x^2+1). Once you see this the first part of the problem is rather straightforward. That is, if a>0, can you cook up an isomorphism

R[x]->R[x]

that takes x^2+a to x^2+1? Why doesn’t it work when a<0?

In fact, when a<0, we get

x^2+a=x^2-(-a)=(x-\sqrt{-a})(x+\sqrt{-a}).

Now use the Chinese remainder theorem.

## One Comment

Thanks a bunch professor Kim – you’re a star. Your time and support is very much appreciated. Really wish I knew about this blog before hand it would have been very useful. TC