# Tag Info

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\$ ...