# Tag Info

8

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

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 ...

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 ...

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 ...

4

Yes, if $f$ is irreducible, then $5\mid [K:\mathbb{Q}]$ where $K$ is the splitting field of $f$. Hence, the Galois group $G$ of $f$ contains a 5-cycle. Furthermore, if it has only two non-real roots, then $G$ also contains a transposition. The result now follows from the fact that if $G<S_p$ (where $p$ is prime) is a subgroup that contains a $p$-cycle ...

3

I claim that $\dim_K(L_F(K)) = [K:F]$: To see this, choose a basis $\{e_1,e_2,\ldots, e_n\}$ of $K$ over $F$ and let $T_{i,j} \in L_F(K)$ be the operator that maps $$e_i \mapsto e_j \text{ and } e_k \mapsto 0 \text{ if } k\neq i$$ Since $\{T_{i,j}\}$ forms an $F$-basis for $L_F(K)$, it follows that it spans $L_F(K)$ over $K$. Furthermore, for any $i$, ...

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 ...

3

This article may be to your liking. It still uses the concept of Galois groups (it's hard to see how one could avoid that), but does not use the Galois correspondence.

3

There exists a much easier way to conclude this fact. Note that $$[\Bbb{Q}(\sqrt[3]{2}) : \Bbb{Q}] = 3$$ implies that there exist only two subfields of $K=\Bbb{Q}(\sqrt[3]{2})$: these are $K$ and $\Bbb{Q}$. Now, $\Bbb{Q}(\sqrt[3]{2}) \cap \Bbb{Q}(\zeta_3 \sqrt[3]{2})$ is a subfield of $K$, but it cannot be $K$ otherwise you would have that $\zeta_3 \in K$, ...

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, ...

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 ...

2

I don't think there's any way to avoid doing a fairly large amount of listing (given that the final answer you are looking for is a list of $18$ degree $4$ polynomials...). However, you at least can avoid going through all $81$ monic polynomials one by one and trying to factor them all. The following is one strategy you might follow. If a polynomial ...

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 ...

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 ...

2

Let $R$ be a ring. Given a right $R$-module $M$ and a left $R$-module $N$, we can form their tensor product denoted $M\otimes_R N$. If $R$ is a field, then a module over $R$ is simply a vector space over that field. As $\mathbb{Z}$ is not a vector space over $\mathbb{R}$, it is not an $\mathbb{R}$-module, so the expression ...

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 ...

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

The polynomial $f(x)$ in general need not have three distinct real roots, and two non-real roots, in order to be not solvable by radicals. Consider, say, $f(x)=x^5+3x+3$, which is irreducible by Eisenstein. Here $f(x)$ has only one real root, because its derivative is positive everywhere, so that $f$ is an increasing function. Nevertheless the Galois group ...

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 $(\Delta f) H \ \Bbb d V = (\text{div} \ \nabla f) H \ \Bbb d V = \text{div} (H \nabla f) \ \Bbb d V - (\nabla f \cdot \nabla H) \ \Bbb ... 1 Since this field$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}$. Hence it is not a splitting field. 1$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$0$) 1 Yes. It is clear that$(X^3-2)(X^3-3)(X^2-2)$splits over$\mathbb Q(j,\sqrt[3]2,\sqrt[3]3,\sqrt 2)$. On the other hand, any splitting field of$(X^3-2)(X^3-3)(X^2-2)$over$\Bbb{Q}$must contain$\sqrt[3]2,\sqrt[3]3,\sqrt 2$and$\frac{j\sqrt[3]2}{\sqrt[3]2}=j$. 1 I don’t see how to do it without any listing at all. I would do it this way, but I’d be using a fairly primitive symbolic-computation package to help me. First, I’d work over$k=\Bbb F_9$, and find an element$z\in S=\Bbb F_{81}\setminus\Bbb F_9$. Then I would do a listing of the elements of$S$, namely all$a+bz$with$a,b\in k$but$a\ne0$. Then I’d find ... 1 In characteristic$2$,$\alpha$is a double root of$X^2-a$cause$(X-\alpha)^2=X^2-\color{red}2\alpha X+\alpha^2=X^2-\alpha^2=X^2-a$. In characteristic$\ne 2$, the polynomial$f(X)=X^2-a$has formal derivative$f'(X)=2X$, which has no root in common with$f$(as$f'(\alpha)=2\alpha\ne0$), hence$f$has no multiple roots. (The fact that$f'$is identically ... 1 No. The simplest counter-example is$t=\sqrt[3]{2}$. 1 In fact, it is an integral domain: the polynomial$f^p-a$is irreducible in$k[x,y]$. In order to show this we use the Eisenstein's criterion. Note that$f^p-a=\alpha^px^p+\beta^py^p-a$. Multiplying by$\alpha^{-p}$doesn't change anything, so we may instead consider the polynomial$x^p+b^py^p-c$,$c\notin k^p$. Now note that the polynomial$b^py^p-c\$ is ...

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