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6

Such a field is called Formally Real.


5

There do not exist such polynomials which are in addition monic, and I would be surprised if this restriction turned out to be fundamental. Assume that $f, g$ is a monic counterexample. In particular, $\gcd(f(x), g(x)) = 1$ in $\mathbb{Z}[x]$. If $f$ has repeated roots we can replace it with $\frac{f}{\gcd(f, f')}$ and similarly for $g$, so we may assume ...


4

Let's get concrete for a moment. Consider the field automorphism: $$f : \mathbb{C} \rightarrow \mathbb{C}\,\,\hbox{where}\,\,f(a + ib) = a - ib$$ This is commonly known as complex conjugation. (It's obvious that it's an invertible and that the domain and codomain are the same, and you can verify for yourself that it is a homomorphism if you want.) If $z ...


4

Hint: The roots of the polynomials $X^p-X$ and $(X+1)(X+2)\dots(X+p)$ areā€¦


4

Let us add the restriction that $f$ is irreducible, since that is the hardest case to consider. The forward direction is simple: if $[E:K] \geq 60$, then Lagrange's theorem would imply that $Aut(K/\mathbb{Q}) \cong A_5$ or $S_5$, which is no good as you've noted. So let's focus on the converse. If $[E:K] < 60$, then $Aut(K/\mathbb{Q})$ is a ...


3

Galois theory is the theory of the duality between profinite groups associated to fields and closed subgroups which arise as dual to field extensions of the original field. It's about the algebra of polynomials over a field and how that helps to understand other fields constructed algebraically from the original field, i.e. from roots of polynomials over a ...


3

If $S$ is any finite set of nonunits of a commutative ring $R$, then $1+\prod S$ is not divisible by any element of $S$. Thus if $R$ is atomic, this implies the set $S$ fails to contain all irreducibles. Specializing, if $\frak P$ is any set of irreducibles (say, of a certain type, like "degree one polynomials") and $S$ any finite subset of $\frak P$, then ...


3

The universal property is that $K(\alpha)$ is the universal (initial) extension of $K$ together with a root of the minimal polynomial of $\alpha$. The algebraic closure has the property that every polynomial has a root, so it is in particular such an extension. (One might think it's the algebraic closure that's supposed to have a universal property here, but ...


2

Since they're Galois, you know that $$\text{Gal}(k(\alpha,\beta)/k)\cong\text{Gal}(k(\alpha)/k)\times\text{Gal}(k(\beta)/k).$$ But then, after the identification you can see that you can produce $[k(\alpha):k][k(\beta):k]$ different automorphisms of $\alpha+\beta$, by acting just on $\alpha$ while fixing $\beta$ from the subgroup ...


2

I disagree with the other two answers. They tell you what the author meant, however this is not necessarily the meaning of "fixes" in mathematics. The author is imprecise. If an automorphism $\sigma$ fixes a subfield $S$ it could mean two things. It fixes the elements of the subfield, $\sigma(s)=s$ for all $s\in S$. That is, it fixes the subfield ...


1

Reed-Solomon Codes and Their Applications Stephen B. Wicker (Editor), Vijay K. Bhargava (Editor) ISBN: 978-0-7803-5391-6 336 pages October 1999, Wiley-IEEE Press http://www.wiley.com/WileyCDA/WileyTitle/productCd-0780353919,miniSiteCd-IEEE2.html


1

Basically, if $K$ is a field, $f: K \to K$ is a field automorphism then $f$ fixes some subfield $F$ of $K$ iff $$ f|_{F} = id_F $$ i.e. $$ \forall x \in F, f(x) = x $$


1

Recall $K(\alpha)=K[x]/(f)$ and that polynomial algebras have a universal property (freely adjoin an element) and quotient rings have a universal property (kill elements in the ideal). It follows that for any $K$-algebra $R$ we have a bijection between $K$-algebra homomorphisms $K(\alpha) \to R$ and roots of $f$ in $R$. And yes, this is a universal property, ...


1

"Early in the development of the subject it was noticed that $\Bbb{Z}$ has many properties in common with $A=\Bbb{F}[T]$, the ring of polynomials over a finite field. Both rings are principal ideal domains, both have the property that the residue class ring of any non-zero ideal is finite, bothe have infinitely many prime elements, and both rings have ...


1

Promoting comments to an answer. If you pick $K=\Bbb{Q}$, $L_1=K(\root3\of2)$, $L_2=K(\omega\root3\of2)$. Neither of those is Galois, but $L_1L_2$ is the splitting field of $x^3-2$ over $K$. Here the answer depends on the exact meaning of the question. If NOT applies to "solvable and Galois", then the counterexample in (1) applies, because neiter $L_1$ nor ...


1

Pinter in "A book of abstract algebra" says it thus: Thus, we are led to the modern notion of algebraic structure. An algebraic structure is understood to be an arbitrary set, with one or more operations defined on it. And algebra, then, is defined to be the study of algebraic structures. It is important that we be awakened to the full generality of the ...



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