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4

Eoin (+1) already showed that $h$ is invariant under $\phi$. By induction on $n$ we see that $\phi^n(r(x))=r(x+na)$ for all $n\in\Bbb{N}$ and all $r(x)\in k(x)$. From this it follows that $\phi$ is of order $p$. So $\phi$ generates a subgroup $G\le Aut(k(x))$ of order $p$. By basic facts of Galois theory the fixed field $K=\operatorname{Inv}(G)$ satisfies ...


3

You can always do the following. It requires a tiny bit of algebraic number theory. Assume that the polynomial has integer coefficients, so $$ p(x)=(x-r_1)\cdots(x-r_n)\in\Bbb{Z}[x],\qquad(*) $$ where $r_i\in\Bbb{C}$ are the zeros. Let $K$ be the splitting field of $p(x)$ over $\Bbb{Q}$. Then $[K:\Bbb{Q}]<\infty$. Let $\mathcal{O}_K$ be the ring of ...


3

In general, not directly. Just take the example of $x^3 - 2$, whose roots are $\sqrt[3]2$, $\rho \sqrt[3]2$ and $\rho^2 \sqrt[3]2$ where $\rho$ is a third root of unity (i.e. $\rho^3 = 1$ and $\rho \neq 1$). These three are not even linearly independent over $\mathbb Q$ since their sum is zero : $$ \sqrt[3]2 + \rho \sqrt[3]2 + \rho^2 \sqrt[3]2 = (\rho^2 + ...


2

You can get a complete answer by checking out the related answer that @JyrkiLahtonen mentions in his comment. But I see this as not so much an algebraic question, as one in complex variable theory and geometry. Do you know the relationship between $2\times2$ complex matrices and fractional-linear transformations of the extended complex plane, also known as ...


2

This looks similar enough to the pair-based definition of the complex numbers that it looks promising to explore how far that analogy will take us. First, it is easy to see that multiplying any pair by $(a,0)$ simply scales both of its elements by $a$. If we identify $(a,0)$ with the usual real numbers (that is sort of handwavy but bear with me for a ...


2

We can show that $h$ is fixed by this mapping by direct computation: $$\varphi(h)=h(x+a)=(x+a)^p-a^{p-1}(x+a)=x^p+a^p-a^{p-1}x-a^p=h$$ Since $k\subset k(x)$ is fixed (the map $f(x)=c=f(x+a)$ for every $a$), we now have a fixed field $k(h)\subset k(x)$ by the above computation and noting that all elements of $k(h)$ are rational functions in the "variable" ...


2

Because subfield and extension feild always have the same characteristic? After all the characteristic is obtained from the prime field, i.e., the smallest subfield (which is the same for subfield and extension field, as it it is loosely speaking just generated from $0$ and $1$), so here $\mathbb Q$. Another definition: Consider the kernel of the additive ...


1

What about binary counting? You can define the addition as if you are adding two numbers modulo $2^n$ using binary representation. Note that $V$ has $2^n$ elements, and $GF(2)$ has $2$ elements. Therefore, the vector space that you are creating will have dimension $n$. Thus, you will end up with $\{0,1\}^n$. Suppose we have two vectors, $\mathbf{u}$ and ...


1

Maybe you’ve forgotten the definition of addition and multiplication in the factor ring (quotient ring) $k[x]/(p)$. The elements are cosets $q+(p)$, where $q$ may be any element of $k[x]$, i.e. any polynomial. You may interpret $q+(p)$ as the set of all things of the form $q$ plus an element of the ideal. Now, you must remember that a coset does not have a ...


1

More properly, $\alpha ^2 +<p(x)>=2+<p(x)>.$. This can be seen by $\alpha^2-2=0$ So, identifying $\alpha=x$, we have $\alpha^2=2$ MOD $p(x)$ means $x^2-2\in<(p(x))>$ which is trivially the 0. $x+<p(x)>$ means the coset of the ideal $<(p(x))>$ formed by taking any polynimal $q(x)$ and then polynomials of the form ...


1

Let $x=a+b\sqrt{5}$, $y=c+d\sqrt{5}$, where $a,b,c,d\in\mathbb{Q}$. Then $$(a+b\sqrt{5})(c+d\sqrt{5})=19,$$ then $$ac+5bd=19,$$ $$ad+bc=0.$$ There are $a,b,c,d\in\mathbb{Z}$ that are solutions: $(a,b,c,d)=(1,2,-1,2)$, $(a,b,c,d)=(8,3,8,-3)$. So, examples: $$x=1+2\sqrt{5}, y=-1+2\sqrt{5};$$ $$x=8+3\sqrt{5}, y=8-3\sqrt{5}.$$ And other kind of ...


1

By unique factorization, the fixed field of the automorphism is the field of fractions of the fixed ring, since assuming the denominator and numerator are coprime your automorphism permutes the irreducible factors appearing in them. To compute the fixed ring use induction on the degree and polynomial division.


1

For odd primes $p$, attempt to solve $X^2 - X - 1 = 0$ using the quadratic formula. Your polynomial is irreducible iff it has no root iff $5$ is not a square mod $p$. I'll leave the $p=2$ case to you.


1

You have two linear equations in $c$ and $d$. You know the right-hand side is $(1,0)$.


1

However for a finite Galois extension with Galois group $G$, you the normal basis theorem: there exists an element $x\in E$ such that its orbit $G\cdot x$ under the action of $G$ is a basis of $E$.


1

A field is a set of numbers which satisfy some calculation rules. For the field the following rules hold: Addition rules Addition has the neutral element 0 Addition has an inverse (adding the negative part to a number) Addition is associative Multiplication rules Multiplication has the neutral element 1 Multiplication is always invertible (by ...


1

Note the following theorem of Jacobson (see, e.g., T.Y. Lam, A First Course in Noncommutative Rings, Theorem 12.10). Theorem Let $A$ be a ring such that, for any $x \in A$ there exists an integer $n(x) > 1$ such that $x^{n(x)}=x$. Then $A$ is commutative. Now let $A$ be a ring satisfying your property. Since $x^5=x$ for all $x \in A$, the ring is ...


1

As an example, let's check distributivity. $\begin{align*} a\boxdot(c\boxplus d)&=a\boxdot(c+d-cd)\\ &=1-t^{\log_t(1-a)\log_t(1-(c+d-cd))}\\ \end{align*}$ $\begin{align*} (a\boxdot c)\boxplus (a\boxdot d)&=(1-t^{\log_t(1-a)\log_t(1-c)})\boxplus (1-t^{\log_t(1-a)\log_t(1-d))})\\ &= ...



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