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0

Suppose $Q(a)$ has degree four and a quadratic extension gives the splitting field $L$. Then the field $L$ has a Galois group which is a transitive subgroup of $S_4$ so it isomorphic to the dihedral group $D_4$ of order 8.


0

Multiantenna modulation uses lattices in number fields (actually division algebras over a number field) as signal constellations. See here or here.


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


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 ...


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 ...


4

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


0

This is a simple and elementary proof, I think. Assume that $K(x,y)=K(t)$ for some $t\in K(x,y)$. Since $x\in K(t)$, there exist (coprime and nonzero) polynomials $u(z),v(z)\in K[z]$ such that $x=\frac{u(t)}{v(t)}$. Consider now the polynomial $f(z)=xv(z)-u(z)\in K(x)[z]$: clearly it is not $0$ and $f(t)=0$, which means that $t$ is algebraic over $K(x)$, ...


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

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. $$ \forall x \in F, f(x) = x $$


3

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 ...


0

Defining separability the way you propose implies that separability and being separably generated are the same thing. Because $K/k$ is a subexetension of $K/k$. (Of course you could require, that one only considers proper subextensions -- but on what basis?) Actually I don't know a convincing simple reason why one wants to have two notions: separability and ...


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 ...


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, ...


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 ...


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 ...


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 ...


6

Such a field is called Formally Real.


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 ...


0

If $F$ is an infinite field and $n<|F|$ as cardinal numbers, then the cardinality of any vector space considered over $F$ of dimension $n$ is equal to the size of $F$. In particular, since $\Bbb C\cong\Bbb R^2$ as vector spaces considered over $\Bbb R$ and $2<|\Bbb R|={\frak c}$, we know $\Bbb C$ and $\Bbb R$ are equinumerous. Given a field $L$ and a ...


1

Hint: Show that the coset of $x-1$ is a zero divisor in the latter ring. Hence you can conclude that it is not a field.


2

$\mathbb{Q}[x]/(x^2-1)$ is not a field. Its a ring with additive group $\mathbb{Q}\oplus\mathbb{Q}$.


4

If $a\in F\setminus K$, then $K(a)$ is of index at least two over $K$. If it is actually already F, just pick b,c at random. Otherwise, Picking $b\in F\setminus K(a)$ makes $K(a,b)$ of degree at least 4. The last steps are exactly the same: can you carry it out without me spelling it out? The easy generalization is that an extension of degree having n ...


10

The dimension is countably infinite. Here is a nice exercise: let $k$ be any field. Then $k(x)$ has a basis consisting of the functions $1, x, x^2, ...$ together with the functions $\frac{1}{f(x)^k}$, where $f(x) \in k[x]$ is monic and irreducible. This is essentially partial fraction decomposition. (Note that the condition that $\mathbb{Q}(e) \cong ...


10

As pointed out in Adam's comment, the dimension will be infinite. The reason is as follows: suppose that the dimension was finite, say $n$. Then any $n+1$ elements of the field must be linearly dependent over $\Bbb Q$; in particular, $$1\,,\ e\,,\ e^2\,,\ e^3\,,\ldots,\ e^n$$ are linearly dependent; so we have $$a_0+a_1e+a_2e^2+a_3e^3+\cdots+a_ne^n=0$$ for ...


1

I suggest you'd rather see this adjunction as a reflector-inclusion pair. $$Q:\mathbf{Dom_m} \dashv \mathbf{Field}:i$$ The forgetful functor is actually a full inclusion $i$ into the category of integral domains and monomorphisms, while the left adjoint $Q$ is the reflector "field of quotients". As is usual with full subcategories, you can think of the ...


3

Yes chose $L$ and $M$ to be two fields with the same infinite transcendence degree.


3

The other posters did a good job explaining what abstract algebra is, so I'll try to help you understand groups. You can think of a group, ring, field etc. as being a set with a certain structure attached to it. The intuition usually comes from concrete examples so I'll include a few. (Disclaimer: This will not be rigorous in the slightest. I'm shooting for ...


4

When you first came to learn maths, what did it show you? For instance, we wrote a lot of equations in terms of x,y,z and so on, but essentially, what they had common is that they were representing some unknown numbers. Abstract algebra is a bit more broad concept: here (in intuitive sense) letters represent pretty much anything, rather than numbers. I ...


23

We learn math with numbers early on. We learn how to apply operations to numbers to get new numbers. We learn rules, and consequences of those rules. All of that is pretty straightforward. But, the real numbers are not the only things we might want to examine in detail. The properties of how elements interact under operations is a more general, abstract ...


16

In "concrete" algebra one works with things like integers, rational numbers, real numbers, complex numbers, matrices, quaternions, permutations, polynomials, geometric transformations (e.g. isometries, similarities, reflections, inversions, projectivities, etc.), etc., subject to operations like addition, multiplication, and composition. In "abstract" ...


6

Mathematics has to do with sets. There is no definition about what the set is, but we all know that set is made up by elements. Many problems in nature can be represented by sets, and the relations of the elements on these sets. The mathematical discipline that studies THE RELATIONS OF ELEMENTS on a given set is called algebra. There are many good properties ...


3

Let $\alpha=a+b\sqrt{k}$ be a zero-divisor then there exists a non-zero $\beta=c+d\sqrt{k}$ such that $\alpha \beta \equiv 0 \pmod{p}$. It is equivalent to $\alpha \bar{\alpha} \beta \bar{\beta} \equiv 0 \pmod{p}$ (where $\bar{\alpha}$ is the conjugate of $\alpha$). This reduces to $$(a^2-kb^2)(c^2-kd^2) \equiv 0 \pmod{p}.$$ Using the prime property, we get ...


1

If $k$ is a perfect square than $ a+ b \sqrt{k} $ is already in $\mathbb Z _ p $ and you're result should follow.


0

A problem in your answer to $(i)$. You don't know that $\phi(l)=l$, because $\phi$ stabilizes $L$ only setwise - not elementwise. You do know that $l'=\phi(l)\in L$ and, consequently, that $\psi(l')=l'$. Using these bits you can surely show that $\phi^{-1}\psi\phi(l)=l$ as required. The argument for the converse part looks fine. You may want to add a ...


3

By Krull's Intersection Theorem we have $⋂_{n≥0}(x,y)^n=0$. I leave you the pleasure to draw the conclusion.


10

More generally, a field with $p^n$ elements contains a subfield with $p^m$ elements iff $m$ divides $n$. In your case, we have $p=2$ and $n=5$, which has no nontrivial divisors. Here is a proof of one direction, the one that concerns the question: If a field $F$ has $p^n$ elements and contains a subfield $K$ with $p^m$ elements, then $F$ is a finite ...


13

Hint: if $K$ is a subfield of $F$, then in particular, $K^*$, the multiplicative group of $K$ must be a subgroup of $F^*$


1

$a = \sqrt{2}, b = - \sqrt{2}$. For a less silly example, $a = \sqrt[3]{4}, b = \sqrt[3]{2}$. Edit: Try $a = i$ and $b$ a root of $x^2 + (1 + i) x + (1 + i)$. This is irreducible by Eisenstein's criterion over $\mathbb{Z}[i]$, although I'm not sure how to show that $\mathbb{Q}(b)$ doesn't contain $\mathbb{Q}(i)$ (and it might not be true).


6

Henning's answer is correct. However, if we let $F$ be the smallest subfield of $K$ and $f(x) = x^2 + 1$ is irreducible in $F$, then your automorphism is legitimate. Here's why: Let $L \subset K$ be its splitting field. Since $L$ is a splitting field of $f$, then it is by definition a Galois extension. At this point, we can consider the automorphism group ...


11

Not necessarily. For example, in the algebraic closure of $\mathbb F_5$, the two roots of $x^2+1$ are $1+1$ and $1+1+1$, and there's obviously no automorphism that interchanges them.


0

Let $H$ be the said set of roots. As $1 \in H$, it is obviously non-empty. If $a, b \in H$, can you show that $ab^{-1} \in H$? Note: Recall that a non-empty subset $H$ of a group is a subgroup if and only if $\forall a, b \in H, ab^{-1} \in H$.


4

Hint: Consider the map $x \mapsto x^n$ on $F^{\times}$. Prove that it is a group homomorphism and consider its kernel.


6

This is not true, one reason being that being an integral domain is not a local property. For a concrete example take $\mathbb{Z}/6\mathbb{Z}$, this is not an integral domain, but if we localise at the set $\{ 1, 3, 5\} = \mathbb{Z}/6\mathbb{Z} \setminus (2)$ we obtain a finite integral domain and hence a field. You can find some details about determining ...


3

Hint. $X^{p^d}-X$ is the product of irreducible monic polynomials whose degree divides $d$. In fact, these are the minimal polynomials of the elements of $\mathbb{F}_{p^d} = \{a \in \overline{\mathbb{F}_p} : a^{p^d}=a\}$. 2nd Hint. $d$ is a prime number. 3rd Hint. Assume that $g$ is a monic irreducible factor of $f$. Show that $g$ has degree $d$. Conclude ...


0

If $\overline{r}\in \mathbb{Z}_p$ and $\overline{r} \neq 0 $ then $\gcd(r,p) = 1 $ and thus there exist $m , n \in \mathbb{Z} $ such that $$mr + np = 1$$ Thus $\overline{m} \cdot \overline{r} = \overline{1} $ in $\mathbb{Z}_p$ , $\overline{r}$ is invertible and so $\mathbb{Z}_p$ is a field. If $n$ is not prime, then $n = ab $ with $a, b \neq n $ and so ...


0

When you take a number mod $n$, you are taking its representative in $\{0,1,\dots, n-1\}$ Use the fact that $a$ (mod n) - $b$ (mod n) $\equiv a-b$ (mod n). (To see this, say $a=m_1n +r_1$ and $b=m_2n+r_2$. Then $r_1+r_2=(a-m_1n)+(b-m_2n)$ so that $r_1+r_2\equiv a+b$ (mod n) ) Hence $a$ (mod n) $= b$ (mod n) $\Longrightarrow$ $a$ (mod n) $-b$ (mod $n$) ...


0

An alternative to the trick in André's answer (+1) would be to use the usual quadratic formula. That works over any field where you can complete the square, which is the case whenever you can divide by two. This in turn is always possible, when $2\neq0$, i.e. when the characteristic is not equal to two. Here we get that $x^2+x+2=0$, iff $$ ...


4

The easiest description of the quadratic extension is that it has the same $25$ elements as yours, with addition defined in the obvious way, and $x^2$ being defined as $-x-2$, that is, $4x+3$. Then $x$ is a solution of your quadratic equation. Once we have defined $x^2$, the product $ax+b)(cx+d)$ can be defined in the "natural" way: Multiply as usual, and ...



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