# Tag Info

## Hot answers tagged field-theory

8

The multiplicative groups ${\Bbb F}_3(X)^\times$ and $\Bbb Q^\times$ are both isomorphic to the sum of $C_2$ (corresponding to $\Bbb F_3^\times$ and $\Bbb Z^\times$) with a free abelian group of countable rank (generated by the primes of $\Bbb Z$ and the irreducible polynomials in $\Bbb F_3[X]$ respectively).

4

In general this takes a bit of tinkering, and more often than not those finite fields are constructed a bit differently. Offering hints/ideas for the following two special cases (both have plenty of details remaining for you!): If $n$ is odd, then $f(a,b)=a^2+ab+b^2$ works. This is because if $f(a,b)=0, a\neq0,$ then it follows that $(b/a)^3=1$. But in a ...

4

You just need to show that $\sum_{\alpha\in F}\alpha^k=0$ for $k=0,1,\dots,q-2$. This is clear for $k=0$ (understanding $0^0$ as $1$). But $\alpha^q-\alpha=0$ for all $\alpha$ so $\alpha^{q-1}-1=0$ for all $\alpha\ne0$, and the result follows from the Newton identities.

2

Nothing wrong with Andres' answer (+1). Offering a possibly technically simpler approach based on the cyclicity of the multiplicative group $F^*=GF(q)\setminus\{0\}$. We know that the group consists of powers of a generator (aka a primitive element) $g$ of order $q-1$. Thus $F^*=\{1,g,g^2,g^3,\ldots,g^{q-2}\}.$ So if $k$ is an exponent, $0< k\le q-2$, ...

2

The ordering on $\mathbb{Q}(x)$ corresponding to $x = \pi$ is archimedean.. The ordering on $\mathbb{Q}(x)$ corresponding to $x = e$ is a different archimedean ordering. The one corresponding to $x$ being positive and infinite is a non-archimedean ordering. Yet these all have the same underlying field. Another example is that we can give two different ...

2

EDIT: The questions were originally stated about the number $a = \sqrt{11} + \sqrt{2}$, so the answer concerns that value of $a$. Notice that $$a^2 = \left( \sqrt{11} + \sqrt{2} \right)^2 = 13 + 2 \sqrt{22}.$$ Hence, $$a^2 - 13 = 2\sqrt{22},$$ and $$\left( a^2 - 13 \right)^2 = 88.$$ Therefore, $$a^4 - 26a^2 + 81 = 0.$$ You can check that the four ...

2

"$L$ is a $K$-vector space" means that $L$ is an abelian group under addition i.e. you can add and subtract members of $L$ and stay in $L$, and that you can multiply elements of $L$ by elements of $K$ (the $K$ elements are thought of as scalars). Multiplication by scalars satisfies the associative law. The addition within $L$ combined with multiplication by ...

1

Closure of$~L$ under multiplication follows in characteristic${}\neq2$ from additive closure and closure under squaring: $xy=\frac12((x+y)^2-x^2-y^2)$. Closure of$~L$ under inverses of nonzero elements follows from the fact that $K/F$ is algebraic: every $x\in L$ satisfies a polynomial equation, that we can assume with nonzero constant terms by dividing out ...

1

It really depends on what you want to study! When you are interested in topological questions, then obviously topological meadows don't include topological fields. In my opinion the natural definition of a topological field should be a (commutative) topological ring $K$ whose underlying ring is also a field such that the map $K^* \to K^*, x \mapsto x^{-1}$ ...

1

If $d=\deg(p)$, the numbers $1,\alpha,\dots,\alpha^{d-1}$ are independent over $F$, as any vanishing linear combination gives us a polynomial of degree less than $d$ where $\alpha$ vanishes. Either this polynomial has all its coefficients $0$, or else $p$ and it have $\alpha$ as a common root and therefore their gcd is nontrivial, contradicting the ...

1

Suppose that $\deg(p) = n$. You have to convince yourself that $F[\alpha]$ is $n$-dimensional, as an $F$-vector space. Can you find a basis? Consider the set $$\left\{ 1, \alpha, \alpha^2, \ldots, \alpha^{n - 1} \right\}.$$ Why is the set linearly independent? Why does it span $F(\alpha)$?

1

A Galois closure of an extension $K/F$ in a fixed algebraic closure $\overline{F}$ is a field which is minimal among all Galois extensions of $F$ containing $K$. One can prove there is a unique such field, which will be the intersection of all Galois extensions of $F$ containing $K$. Suppose $L$ is the Galois closure of $K/F$. Then $L$ must contain all ...

1

1) It's because $A$ is defined to be $k[x]/(q(x))$. Using $\overline{f}$ to denote the residue class in $A$ of an element $f \in A$ you have: $a = \overline{x}$ and so $q(a) = q(\overline{x}) = \overline{q(x)} = \overline 0$. (The critical step you might be missing is $q(\overline x) = \overline{q(x)}$, which holds more generally as $q(\overline f) = ... 1 Every element of$F_D$can be written as$x/y$with$x,y\in D$and$y\ne0$(but not uniquely so, beware!). First prove that$F_\theta(x/y) = \theta(x)/\theta(y)$is a well-defined field homomorphism. Next prove that every field homomorphism$F_D \to K$extending$\theta$must send$x/y$to$\theta(x)/\theta(y)$and so must coincide with$F_\theta$. It's ... 1 Indeed, all the finite field extensions are of the form$\mathbb F_{p^n}$(up to isomorphism), and all these fields have characteristic$p\$. Note that in general, the characteristic of a field is always equal to that of its prime subfield. Another way of looking at this: the only way you can have a morphism between two fields (remember that they are always ...

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