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4

1) If you know that every irreducible polynomial over $\mathbb R$ has degree $1$ or $2$, you immediately conclude that $\mathbb C$ is algebraically closed: Else there would exist a simple algebraic extension $\mathbb C\subsetneq K=\mathbb C(a)$ with $[K/\mathbb C]=\operatorname {deg}_\mathbb C a=d\gt 1$. But then the minimal polynomial $f(X)\in \mathbb ...


3

Since the two polynomials are irreducible (no roots, small degree) you simply need to invoke the uniqueness of finite fields of a given order: Finite field extensions are uniquely characterized by their order. Since both of them are finite fields of order $3^3=27$ they are isomorphic, and in fact are exactly the set of all solutions to $x^{27}-x=0$. If ...


2

If it had a factor that is a linear polynomial in $\mathbb{Q}$ then it would have a rational root. The only possibilities for rational roots are divisors of $4$: $\pm1$, $\pm2$, or $\pm4$. We can check if these are solutions. If it factors as $(x^2+ax+b)(x^2+px+q)=x^4+(a+p)x^3+(b+q+ap)x^2+(bp+aq)x+bq$ So, we get ...


2

Let $\varphi : \mathbf{R}\to \mathbf{R}$ be a morphism of fields. The kernel of $\varphi$ is an ideal of the field $\mathbf{R}$, and a (commutative) field $k$ has only two ideal, $(0)$ and $k$, and the kernel of $\varphi$ can't obviously be equal to whole $\mathbf{R}$, as then we would have $\varphi(1)=0$ and at the same time (as $\varphi$ is a morphism of ...


2

Let us show that every morphism of fields $f:\mathbb R\to \mathbb R$ is the identity, namely $f(r)=r$ for all real $r$. 0) Trivially $f(q)=q$ for all $q\in \mathbb Q$ 1) Notice that $f$ preserves the order relation in $\mathbb R$ : $x\leq y \implies f(x)\leq f(y)$. Indeed if $r\geq 0$ we can write $r=\rho^2$ for some $\rho\in \mathbb R$ and then ...


2

There is no problem. There is just no irreducible polynomial over a finite field of the form $g(x^p)$.


2

$\newcommand{\F}{\mathbf{F}}$I believe you need to consider the following classical example. Let $x, y$ be independent indeterminates over the field $\F$ with $p$ elements, $p$ a prime. Consider the fields $E = \F(y)$ of rational functions over $\F$ in the indeterminate $y$, and its subfield $K = \F(y^{p})$. Then one can prove that the polynomial $$ f(x) = ...


2

Hint: $\left(\sqrt{3+\sqrt{5}}\right) \left(\sqrt{3-\sqrt{5}}\right)=?$


2

For $(2)$, we have that $2011$ is a prime number taking the thousand and eleventh root of unity, that is $\xi \neq 1$. Then $L = \mathbb{Q}[\xi]$ and as $$\xi^{2010}+\xi^{2009}+\ldots +\xi^2 + \xi + 1 = 0$$ and $p(x) = x^{2010}+x^{2009}+\ldots +x^2 + x + 1$ is irreducible* over $\mathbb{Q}$ then $[L:\mathbb{Q}] = 2010$. To find the automorphism notice ...


2

You need more detail in (1) about how you write $z_3$ in terms of $z_1$. The most important point is that $\sqrt{2} \in \mathbb{Q}(z_1)$. In (2), you just need to show that there is an automorphism $\tau$ with $\tau^2 \ne \operatorname{id}$. So you just need to check that $\tau_1(z_3) \ne z_1$. I don't understand your calculation there. To begin with, ...


2

For (1), there's some $m_i$ that works for each $\alpha_i$. It's enough to let $m$ be the largest of these. For $(2)$, assume that $E^{p^{m-1}}k = E$. Apply the Frobenius endomorphism to each side of this equality to obtain $E^{p^m} k^p = E^p$, hence $E^{p^m} k = E^{p^m} k^p k = E^p k = E$.


1

If F is a splitting field over K, then that is normal over K. But the constructed field is not normal over the base field, because x^n-3 is irreducible (by Eisenstein's Criterion); and the field has a root, and does not have all roots of x^n-3. (n is greater than 2)


1

You already know the extension's Galois Group is of order four, and there aren't that many groups of that order. You may also want to use that in fact $$L=\Bbb Q(\sqrt2+\sqrt3)=\Bbb Q(\sqrt2\,,\,\sqrt3)\;$$ From the above, you must be able to deduce your group isn't cyclic.


1

Hint (and actually an answer spoiling all suspense) : induction on $n$.


1

How many real roots does it have? (Use calculus.)


1

First, observe that each of the polynomials $x^3-x+1$ and $x^3-x^2+x+1$ are irreducible over $\mathbb Z_3$. To prove this, we use the following fact. If $\deg p\leq 3$ and $p$ has no roots, then $p$ is irreducible Proof: If $p$ was reducible, there would be non-constant polynomials $f,g$ such that $p=fg$. Then by degree considerations, we have either ...


1

Absolutely nothing wrong with the other answers. But if you want a concrete isomorphism you can find one (in this case) with the following ad hoc trick. Let $\beta$ be a zero of the polynomial $p(x)=x^3-x^2+x+1$. Then $$ (\beta+1)^3-(\beta+1)^2=(\beta^3+1)-(\beta^2+2\beta+1)=p(\beta)-1=-1. $$ So $$ (\beta+1)^3-(\beta+1)^2+1=0. $$ Dividing this equation by ...


1

I think the answer is yes if $X$ has bounded period, for then $X$ is necessarily finite and cyclic. If $X$ has order $d > 1$, choose $\alpha \in F$ of order $d,$ and let $\alpha$ have minimum polynomial $f(x)$ over $K = {\rm GF}(p).$ Then $f(x)$ divides $x^{d}-1$, and the degree of $f(x)$ is $m,$ the smallest positive integer such that $d$ divides ...


1

This is a variant of the dimension theorem about field extensions. Suppose $V$ is finite dimensional over $L$ and $L$ is finite dimensional over $K$. Let $\{v_1,\dots,v_m\}$ be a basis of $V$ over $L$ and $\{l_1,\dots,l_n\}$ be a basis of $L$ over $K$. Then, if $v\in V$, we have $$ v=\sum_{i=1}^m\alpha_iv_i $$ with $\alpha_i\in L$. Therefore we have $$ ...


1

$\xi_n$ is in $K$ iff the group $(\mathbb Z/n\mathbb Z)^*$ (of invertible elements of $\mathbb Z/n\mathbb Z$) is isomorphic to a product of cyclic groups which are of order $\leq 4$ $\ \ \ \ \ \ \ \ \ (*)$ to prove it: one direction: if $L$ is the splitting field of $k$ polynomials of degree $4$ in $\mathbb Q[x]$ then $G=Gal(L/\mathbb Q)$ is a ...


1

I am showing only the non trivial implication, you showed the other. 1) First suppose that $K$ is algebraically closed. 1)a) Let $\mathfrak{m}$ a maximal ideal of $A$. $B = A / \mathfrak{m}$ is a field ($\mathfrak{m}$ is maximal) and is a finite extension of $K$ as $A$ is a finite dimension over $K$. As $K$ is algebraically close, $B$ is of dimension $1$ ...



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