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0

The other answers are so long! One field has a square root of negative one call it $i$ for obvious reasons. The other one doesn't. If $\phi$ is an isomorphism what is $\phi(i)^2$?

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Your definition should read "combindend with a binary operation $+$ and $\cdot$", also it should better say that $+$ turns $F$ into an abelian group. Thereby the difference become more apparent: The set with the operator $+$ is not only a group, it's abelian (or commutative) The set (excluding the zero) with the operator $\cdot$ is also not only a group, ...

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$x^2=x\iff x^2-x=x(x-1)=0$ Since it is a quadratic equation a coefficients in a field, there are just two solutions which are given by the last equality, (i. e. $1$ and $0$)

8

Since $F_5$ is a field with only five elements, it is perhaps simplest to solve the equation by just trying each element.

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Uniquesolution gives a good answer. The answer to your second question "is a vector space over $F$ just a collection of vectors with coordinates in $F$?" is yes. A particular vector space has dimension, which is just the number of different entries in a (column) vector drawn from the space. Two columns of different length have to live in different vector ...

3

The binary operations for a field are not restricted to just $+$ or $\cdot$. Like $*$, the symbols $+$ and $\cdot$ are just generic symbols for arbitrary binary operations. However, a field needs two operations (i.e., the definition should read "$+$ and $\cdot$" instead of "$+$ or $\cdot$", and accordingly the tuple should be $(F,+,\cdot)$ instead of ...

6

The first thing you should bear in mind is that fields have two operations, whereas groups only have one. (In your notes, the word "or" is misleading, as it should really be "and"). These two operations are more often than not called "addition" and "multiplication", so they are customarily denoted by $"+"$ and "$\cdot$", because the most natural examples of ...

1

Your statement follows from Stokes's theorem $\int \limits _{\partial M} \omega = \int \limits _M \Bbb d \omega$ and the fact that $M$ has no boundary (i.e. $\partial M = \emptyset$, so the left-hand side is $0$). Note that $(\Delta f) H \ \Bbb d V = (\text{div} \ \nabla f) H \ \Bbb d V = \text{div} (H \nabla f) \ \Bbb d V - (\nabla f \cdot \nabla H) \ \Bbb ... 2 Not sure how the gradient of a specific function$f$can be interpreted as an operator. The integral of the divergence of a vector field on a compact manifold is zero. Apply this to the vector field$H\nabla f$(scalar times vector). But the divergence of$H\nabla f$is equal to the inner product of$\nabla H$with$\nabla f$plus the product (as scalar ... 4 The answer depends on the cardinality of the underlying field.$K(x)$, as a$K$-vector space, has a basis consisting of the polynomials$1, x, x^2, \dots$together with the rational functions$\frac{x^k}{f(x)^n}$where$f(x)$runs over all monic irreducible polynomials in$K[x]$,$k < \deg f$, and$n$runs over all positive integers. This is a corollary ... 0 From the first part of the proof we know three things:$f$is irreducible (because$\mathbb{Z}_p[x]/\langle f(x)\rangle$is a field)$f$has degree$n$(because the degree is the same as the degree of the extension)$f$has some zero in$E'$(because$E'$is exactly the set of roots of the polynomial$x^{p^n} - x$, of which$f(x)$is a factor). Now, let ... 0 If$k$is a field, then$K\supset k$is a splitting field over$k$if there is a$k$-polynomial$f\in k[X]$such that$K$is the field gotten from$k$by adjoining all roots of$f$. In my world, it does not make sense to say “$K$is a splitting field” unless you mean for$K$to be a splitting field over$\Bbb Q$. To give an explicit example,$\Bbb C$is a ... 0 You could also show that $$\mathbb Z[x]/(2x+3) \to \mathbb Z\left[\frac{1}{2}\right]\\ x \mapsto -\frac{3}{2}$$ is an isomorphism. 2 Hardly an answer, but an easily-seen unit in$\Bbb Q(\sqrt[3]2\,)$is$\alpha-1$in your notation, and it’s hard to imagine another unit closer to the identity (in the unique real embedding of that field), though I’m not skilled enough to say that it’s a fundamental unit. For the full field$K$, which has three inequivalent complex embeddings, two obviously ... 1 Since this field$K$is contained in$\mathbb{R}$, it does not contain one root,$\beta$of the polynomial$x^7 - 5$. There is a homomorphism from$K$to$\mathbb{C}$which sends$\alpha$to$\beta$and fixes$\sqrt{5}$. This homomorphism does not map$K$to itself, so$K$is not a normal extension of$\mathbb{Q}$. Hence it is not a splitting field. 6 No. For example,$\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$is not a normal extension. The minimal polynomial of$\sqrt[3]{2}$is$x^3 - 2$, but$\mathbb Q(\sqrt[3]2)$does not contain$e^{2\pi i/3} \sqrt[3]{2}$and$e^{2\pi i/3} \sqrt[3]{2}$is another root of$x^3 - 2$. Since$x^3 - 2$is the minimal polynomial for$\sqrt[3]{2}$, every normal extension that ... 1 No. The simplest counter-example is$t=\sqrt[3]{2}$. 2 The Galois group of your extension is not$Z_2\times Z_3$. This group is cyclic if order$6$, but as the Galois group of a cubic polynomial, it must be a subgroup of$S_3$, which it isn't. In fact, the Galois group is$S_3$, which is not abelian. This is immediate from the fact that$G$must be a subgroup of$S_3$of order$6$. To see it explicitly, ... 1 You showed that the splitting field could be obtained by adjoining a single root of a quadratic polynomial irreducible over the rationals. So the splitting field has degree$2$over the rationals. The field$\mathbb{Q}(i,\sqrt{3})$indeed has degree$4$over the rationals. But it is not the splitting field of$t^3-1$. The degree argument shows that. If you ... 1 Let$f$be a quadratic polynomial over a field$F$. If$f$has a root in$F$, then also the other one is in$F$, by simple factorization. For the same reason, if$f$has a root in an extension field$K$of$F$, then both roots belong to$K$. In your case, the root$1$can be ignored, because it's rational. When you add either root of$t^2+t+1$to ... 2 The roots are$1, \mathrm j, \mathrm j^2$. Hence the splitting field is$\;\mathbf Q(1,\mathrm j, \mathrm j^2)=\mathbf Q(\mathrm j)$, which has degree$2$. The minimal polynomial of its generator is$x^2+x+1$. This splitting field is contained in$\mathbf Q(\mathrm i,\sqrt 3)$, which is the splitting field of$(x^2-3)(x^2+1)$, but certainly not equal. The ... 1$BA$is obtained multypling the first row for$a_{11}$, second row for$a_{22}$, ... n-th row for$a_{nn}$;$BA$is obtained multypling the first column for$a_{11}$, second column for$a_{22}$, ... n-th column for$a_{nn}$Now it appears clear that$AB=BA$only and only if$B$is diagonal. 3 Adding in details: In$\mathbb Z_2$, every element is its own additive inverse:$0+0=0$,$1+1=0$, so$0=-0$,$1=-1$. Another way of saying this is that$1+1=0$, and it makes this a "Field of Characteristic 2". So,$\forall b\in \mathbb Z_2,b=-b$. Now, subtraction is defined as adding the additive inverse in general, so the way we define subtraction is ... 0 If$k \neq j$, then $$[BA]_{kj} = \sum_{r} [B]_{kr} [A]_{rj} = [B]_{kj} [A]_{jj},$$ and $$[AB]_{kj} = \sum_{r} [A]_{kr} [B]_{rj} = [A]_{kk} [B]_{kj}.$$ And, $$[BA]_{kk} = \sum_{r} [B]_{kr} [A]_{rk} = [B]_{kk} [A]_{kk},$$ and $$[AB]_{kk} = \sum_{r} [A]_{kr} [B]_{rk} = [A]_{kk} [B]_{kk}.$$ Thus, we have $$[AB]_{kk} = [A]_{kk} [B]_{kk} = [BA]_{kk}.$$ Now ... 1 Also one can use the fact that$\mathbb{F}_{p^2}/\mathbb{F}_p$is Galois, in other words is normal, so one can write$\prod (x^2-i)$actually splits in$\mathbb{F}_{p^2}$. So, ans is$\mathbb{Z}_2$4 By the uniqueness of finite fields of a given degree, there is one and only one field of order$p^2$. Since not all elements of$\Bbb F_p$are squares, you know that if$a$is a non-square then $$\Bbb F_p[\sqrt{a}]=\Bbb F_p[x]/(x^2-a)=\Bbb F_{p^2}$$ is an extension of degree$2$. But then since another non-square$b$would also generate an extension ... 2 As Georges Elencwajg said, a$K$-endomorphism of an algebraic extension$F$is indeed an isomorphism. Here is another argument. Let$f:F\rightarrow F$be your$K$-endomorphism. First, note that it is into, because its kernel is an ideal of$F$, and since it is not$F$, it is$\{0\}$. We want to prove it is onto, so let$a\in F$. Because$a$is algebraic ... 1 Yes, the result holds also for endomorphisms$f:F\to F$of infinite algebraic extensions$F/K$. Proof: Let$a\in F$be an arbitrary element and let's show that$a$is in the image of$f$. Consider the field$N=K[a_1,\dots, a_r]$obtained by adjoining to$K$the roots$a=a_1,\dots, a_r$in$F$of the minimal polynomial$Irr(a,K,X)$of$a$over$K$. Notice ... 1 In the field$\Bbb F_2[x]/(x^3+x+1)$, one root of$X^3+X+1$is$\alpha=x$, or more precisely the coset of$x$with respect to the ideal$(x^3+x+1)$. You should be able to check quickly that the other two roots are$\alpha^2$and$\alpha^4$. The reason for this is that$z\mapsto z^2$is an automorphism of every finite field of characteristic two, and is ... 1 Another argument uses the inclusions$\Bbb Q\subset\Bbb Q(\sqrt2)\subset\Bbb Q(\sqrt[6]2)$and the “well-known” fact that$\Bbb Z[\sqrt2]$is a principal ideal domain and therefore has unique factorization. Irreducibility of$X^3-\sqrt2$over the quadratic field follows from the fact that the polynomial has no roots in$\Bbb Q(\sqrt2)$, and this follows from ... 1 It's easier if you write$[3x-2]=3r-2$, where$r^2=2$. Then $$\frac{1}{3r-2}=\frac{3r+2}{(3r-2)(3r+2)}=\frac{3r+2}{18-4}= \frac{3}{14}r+\frac{1}{7}$$ More generally, any element of your field can be written in a unique way as$a+br$, with rational$a$and$b, not both zero. Then \begin{align} (a+br)^{-1} &=((a+br)(a-br))^{-1}(a+br)\\[6px] ... 1 You do to complicate: $$(3x-2)(ax+b)=3ax^2+(3b-2a)x-2b$$ and thus, $$(3x-2)(ax+b)=k(x^2-2)+1\iff(3a-k)x^2+(3b-2a)x-2(k-b)-1=0\iff...$$ 1 Suppose\alpha\in K$. Then there is some nonzero polynomial$p(x)\in F[x]$such that$p(\alpha)=0$. Clearing denominators, we get a polynomial$q(x)\in A[x]$such that$q(\alpha)=0$. Now if$q(x)=cx^n+dx^{n-1}+ex^{n-2}+\dots$, we see that $$c^{n-1}q(x)=(cx)^n+d(cx)^{n-1}+ce(cx)^{n-2}+\dots=r(cx)$$ for some monic polynomial$r(x)\in K[x]$. Thus ... 3 It never has a left adjoint. The argument in this answer will work with essentially no modifications. 3 The field is given by$K=\mathbb{Q}(i,\sqrt{2})$. We can construct it from$\mathbb{Q}$by a quadratic extension$L=\mathbb{Q}(\sqrt{2})$with minimal polynomial$x^2-2$, and then again a quadratic extension$K=L(i)$with minimal polynomial$x^2+1$. Since clearly$i\notin\mathbb{Q}(\sqrt{2})$, we have $$[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2}]=2.$$ Now ... 3 Yes, your argument is correct: it is enough to prove that$t=\sqrt[3]{2}$has no square root in$\mathbb{Q}(t)$. To show this directly, we can calculate$(a+bt+ct^2)^2 = (a^2 + 4bc) + (2c^2 + 2ab)t + (b^2 + 2ac)t^2$. This gives us three equations to be solved in rationals: $$a^2 + 4bc = 0$$ $$2c^2 + 2ab = 1$$ $$b^2 + 2ac = 0$$ If$a=0$or$b=0$, then the ... 3 This isn't even well-defined, because the function$\ln$is not a bijection (or even always well-defined!) so$\exp$(and hence$+$) are not always well-defined. When$a^2+b^2=1$, the first coordinate of$\ln(a,b)$is$0$, so it does not give an element of the set; maybe you are saying that$\ln(a,b)=0$for such$(a,b)$(with$0$being a special element ... 1 Here's an answer that uses a bit more theory. First some generalities. Suppose$F$is a field,$f \in F[x]$is irreducible and$K = \frac{F[x]}{(f(x))} = F(\theta)$is the extension of$F$obtained by adjoining a root$\theta$of$f$(which in your notation is$[x]$, the image of$x$in the quotient). Given$\sigma \in \text{Aut}(K/F)$, then it follows by ... 1 The splitting field is, by definition or a brief argument, the field generated by the roots of the polynomial(s). Thus when$a_1, a_2, a_3, a_4$are your roots (possibly some of them equal), then the splitting field is$\mathbb{Q}[a_1, a_2, a_3, a_4]$. You, then can simplify the description, for example by removing rational summands, or multiplying by ... 1 Hint: The roots of the quadratics are all in$\mathbb{Q}(\sqrt{-3})$. 1 If$f$is reducible, say$f(x)=h(x)g(x)$where$h(x)$is irreducible, then you can use a similar argument when adding a root of$h$to the field. Note that it is just fine even if$f(x)$has a root (in this case,$h(x)=x-a$and adding$a$simply wont extend the field). If you could show how to add a root to a field, without assuming a containing splitting ... 1 Yet a third argument: We can still apply Eisenstein over$\mathbb{Z}[i]$, we just need to check that$(3)$is a prime ideal in$\mathbb{Z}[i]$. One quick way to do this is to compute$\mathbb{Z}[i]/(3) \cong \mathbb{Z}[X]/(3,X^2+1) \cong (\mathbb{Z}/3\mathbb{Z})[X]/(X^2+1)$, and verify that$X^2+1$is irreducible in$\mathbb{Z}/3\mathbb{Z}$. 1 Suppose$p(x)$is reducible over$Q(i)$. It is cubic and so it must have a root in$Q(i)$. Then the degree of$\min(3^{1/3},Q(i))$is$1$.$Q(i,3^{1/3})$is the field extension of$Q(i)$containing the root of$p(x)$. Now$[Q(i,3^{1/3}):Q]=[Q(i)(3^{1/3}):Q(i)][Q(i):Q]$and also$[Q(i,3^{1/3}):Q]=[Q(3^{1/3})(i):Q(3^{1/3})][Q(3^{1/3}):Q]$. So we have by ... 1 Here is a crude argument. If our cubic is reducible over$\mathbb{Q}(i)$, it has a root in that field. But we can easily write down the roots, and show none is in$\mathbb{Q}(i)$. That basically comes down to the irrationality of$\sqrt[3]{3}$. Remark: After one has a few tools, "nicer" arguments are available. If$x^3-3$is reducible over$\mathbb{Q}(i)$, ... 0 Yes. For instance, a polynomial with infinitely many roots must be$0$, so you can take$f$to be any function which is$0$at infinitely many points but not identically$0$. In fact, almost all functions$F\to F$are not given by polynomials, in a way that you can make precise using cardinalities. You can show that the cardinality of the set of ... 2 To expand on what I said in the comments: Your argument for$3$(and also your argument in the comments) works in this particular case, because the degree of$X^2 + X+1$is small. Your claim that the minimal polynomial of$j$over$\mathbb Q(\sqrt[3]2)$is the same as the minimal polynomial of$j$over$\mathbb Q$because$j\notin \mathbb Q$works in this ... 3 If the minimal polynomial of$x$is the same as the minimal polynomial for$y$then$z$is a root of this polynomial and transitivity holds. But a minimal polynomial is irreducible (by minimality - easily seen because we are over a field so there are no zero divisors) so this is straightforward. See comments: this depends actually on an assumption about ... 0 You have written the answer. The point is that$X^p-t$has a unique root thus its splitting field is normal by definition. 1 The conjugates of$x$are the roots of$\text{Irr}(x,K)$, and therefore the coefficients of the minimal polynomial are$s_i(x_1,\dots,x_n)$, where$s_i$is the$i$th symmetric polynomial. See Vieta's formulas. 1 A very direct way to see that$K[X]/(f)$is an algebraic extension of$K$is as follows: First of all$K[X]/(f)$is a field. Indeed,$K[X]$is a ring and$(f)$is a maximal ideal in it (because$f$is irreducible), hence the quotient is a field$K[X]/(f)$is an extension of$K$. Indeed, the canonical map$K\to K[X]/(f)$,$a\mapsto a+(f)\$ is a ring ...

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