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0

Denote $\alpha = \sqrt[3]{2}+ \sqrt{5}$. Let $\zeta = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$\ $ The product \begin{eqnarray} P(X) =\prod_{0\le k \le 2, 0 \le l \le 1} ( X - (\sqrt[3]{2}\cdot \zeta ^k + \sqrt{5}\cdot(-1)^l)\ ) \end{eqnarray} is a polynomial with integer coefficients. We calculate: \begin{eqnarray} P(X) = X^6- 15 X^4 - 4 X^3 + 75 X^2- ...


4

Consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{2}+\sqrt{5}) \subset \mathbb{Q}(\sqrt[3]{2},\sqrt{5})$. We know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{5}):\mathbb{Q}]=6$ so the only possibilities for $[\mathbb{Q}(\sqrt[3]{2}+\sqrt{5}):\mathbb{Q}]$ are 2, 3, and 6. It does not take long to verify that $\sqrt[3]{2}+\sqrt{5}$ is not a root of ...


2

Note that $\mathbb{Q}[\sqrt[3]{2} + \sqrt{5}] \subset \mathbb{Q}[\sqrt[3]{2}, \sqrt{5}]$. Since $\mathbb{Q}[\sqrt[3]{2}, \sqrt{5}]$ is a degree $6$ extension over $\mathbb{Q}$, then it must be the case that $[\mathbb{Q}[\sqrt[3]{2} + \sqrt{5}]:\mathbb{Q}] = 2, 3, \text{or } 6$. Could $\sqrt[3]{2} + \sqrt{5}$ be a root of a quadratic? Of a cubic? If the ...


1

A set on which the operations of arithmetic, subtraction, multiplication and division hold.


1

What you constructed is a rupture field for the irreducible polynomial; it has dimension $n=\deg p$ as vector space over the ground field $F$. A minimal extension field where $p$ has $n$ roots (if we suppose it to be separable) is called a splitting field for$~p$; it's dimension is some divisor of $n!$, so it can be considerably larger than a rupture field. ...


1

is most definitely not true. Consider $\mathbb{Q}[X]/(X^2+1)$ for instance. $\pi(-X)$ is also a solution. In fact if the extension is 'galois' then it's most definitely false (if deg $p(X)>1 $ anyways. Actually it's part of definition.) If the extension is Galois then all roots of p(x) must exist in $L$ (sorry I can't remember if that extension is ...


4

As explained in this math.SE answer and stated in the comments by mercio, for an irreducible polynomial of degree $n$ this condition is equivalent to the Galois group failing to contain an $n$-cycle. There are many transitive subgroups of $S_n$ failing to contain an $n$-cycle and hence (up to solving the inverse Galois problem, also as stated in the comments ...


2

An inclusion is obvious: $$\mathbb{Q}(\sqrt{2}(1+i))\subseteq \mathbb{Q}(\sqrt{2},i\sqrt{2})$$ and the reverse inclusion follows from: $$\left(\sqrt{2}(1+i)\right)^2 = 4i,\qquad \left(\sqrt{2}(1+i)\right)-i\left(\sqrt{2}(1+i)\right)=2\sqrt{2}.$$


2

It is not the splitting field since it is not Galois. Such fields contain all the conjugate roots, $\zeta_n\sqrt[n]{3}$


2

Hint: If $K$ is a splitting field of some polynomial over $F$, and if some other irreducible polynomial over $F$ has a zero in $K$, then this other polynomial splits over $K$.


2

It is isomorphic to $\Bbb Q[x]/(x^2-2)$, since the polynomial has degree $2$, the vector space dimension is $2$. If you don't know this fact, it's easy to see by since a polynomial has reduction modulo $(x^2-2)$ given by $$p(x)=\sum_{n=0}^{N_1} a_{2n} x^{2n}+\sum_{n=0}^{N_2}a_{2n+1}x^{2n+1}$$ $$\overline{p(x)}=\sum_{n=0}^{N_1} a_{2n} ...


3

Show that $\{1,\sqrt{2}\}$ is a basis.


0

The main thing you want to show is that multiplicative inverses are unique: that is, choose $x \neq 0$ in your field. Then the field axioms guarantee an element $y$ such that $xy = 1$, but you can conclude more strongly that there is a unique such $y$. Given that, it isn't so bad. Remember that the inverse of $x$ is defined to be the (now unique) element ...


0

For multiplicative inverses in a ring/field, we only look at the elements of the ring/field that are NOT the 0 element (The additive identity). You probably are being asked in the first inverse question to show whether they have additive inverses though, because that is a necessary quality for a field to have. Of your sets which are closed under addition, ...


0

The word "dimension" isn't usually used for the size of a minimal generating set for an algebra relative to some subalgebra (i.e. $\{1,x\}$ is a generating set for $\mathbb{k}(x)$ as an algebra over $\mathbb{k}$, but I wouldn't call it a "basis"). "Dimension" in the context of an extension field means vector space dimension (unless the author is doing ...


2

The mapping $N(a)=a^{p+1}$ is known as the norm map from $\Bbb{F}_{p^2}$ to $\Bbb{F}_p$. It is a homomorphism of multiplicative groups, but, as you observed, does not respect addition. Its mapping properties can be most easily deduced as follows. The multiplicative group of $\Bbb{F}_q$ (resp. $\Bbb{F}_p$) is cyclic of order $q-1=(p-1)(p+1)$ (resp. cyclic of ...


2

$(p=3) \implies f(-1)=f(+1)$ but $-1 \neq +1$ Hence the map is not an automorphism. your verification of the property of addition is not needed to verify the map.


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From this link, it seems to me that you have misunderstood. The author is explaining the idea of a splitting field. A splitting field for $X^3 - 2$, by definition, should have all the roots of $X^3 - 2$. Certainly $\sqrt[3]{2}$ is one such root and it should belong to the splitting field. But $\sqrt[3]{2}\omega$ and $\sqrt[3]{2}\omega^2$ are also roots of ...


3

There are three complex roots $$\sqrt[3]{2}, \sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}, \sqrt[3]{2}\frac{-1-\sqrt{-3}}{2}$$ only $\sqrt[3]{2}$ is real.


0

It sounds like you have ascertained that the splitting field is $F=\Bbb{Q}(\root7\of{13},\zeta_7)$ where $\zeta_7=e^{2\pi i/7}$. Let us denote the key intermediate fields by $L=\Bbb{Q}(\root7\of{13})$ and $K=\Bbb{Q}(\zeta_7)$. We have, indeed, $[F:\Bbb{Q}]=42$, because $[L:\Bbb{Q}]=7$ (Eisenstein), $[K:\Bbb{Q}]=6$ (cyclotomic) and $\gcd(7,6)=1$. An ...


2

Yes. Sketch: $k[X, Y]$ is a UFD, so it makes sense to ask that a fraction $\frac{p(x, y)}{q(x, y)} \in k(X, Y)$ be in lowest terms. The first condition implies that $q(x, y)$ is a polynomial in $x$ only and the second condition implies that $q(x, y)$ is a polynomial in $y$ only.


4

I assume $p$ is an odd prime. The order of $(\mathbb{Z} / p^n)^*$ is $(p-1)p^{n-1}$. You can decompose it as the direct sum of the subgroup of $(p-1)$-th roots of unity and the group of elements that are $1 \bmod p$. We can define the logarithm function on the latter group as $$ \log(1 + x) \equiv \sum_{i=1}^{\infty} (-1)^{i+1} \frac{x^i}{i} \pmod{p^n}$$ ...


4

You can do it if and only if the torsion subgroup is a subgroup of $\mathbb{Q} / \mathbb{Z}$: i.e. it has at most one subgroup of order $n$ for every finite $n$. If this is the case, then you can write the group as a product $T \times F$ where $T$ is torsion and $F$ is torsion-free. Let $F \otimes \mathbb{Q} \cong \mathbb{Q}^\alpha$. Then if $K$ is an ...


1

The imaginary number $i$ is defined to be the square root of $-1$. This means that $i^2=-1$, and therefore $$zw=ac+adi+bci+bdi^2=ac+adi+bci+bd(-1)=(ac−bd)+(ad+bc)i.$$ For a complex number $x=p+qi$, where $p$ and $q$ are real numbers, it is common to write $\text{Re}(x)=p$ for the real part of $x$ and $\text{Im}(x)=q$ for the imaginary part of $x$. For any ...


2

Yes, the multiplicative inverse of $z$ is $1/z$ (for nonzero $z$). But the point is, what exactly is the complex number $1/z$? Remember, a complex number is usually expressed in the form $$(\textrm{real number}) +(\textrm{real number})i$$ Your instructor is trying to find what the two real numbers (the real and imaginary components of $1/z$) are--that's ...


1

First, remember that $i^{2}=-1$ , be definition. Second, she wanted to find the real numbers $c$ and $d$ such thath $w=\frac{1}{z}$, and that is why she expressed them in terms of $a$ and $b$ (an explicit expression of $w$.)


1

Yes, the multiplicative inverse of $z$ is $1/z$. Your teacher is finding an explicit value for $1/z$. This both proves that it exists (she has displayed it) and is useful in the future for computation.


1

Suppose $K$ has dimension $n$ over $F$, then the $n+1$ elements $1,a,a^2 \dots a^n$ are linearly dependent over $F$. An explicit dependence gives a polynomial over $F$ which is satisfied by $a$.


3

Nope! Any finite multiplicative group of a field is cyclic. See here, for example http://www.math.upenn.edu/~ted/203S06/References/multsg.pdf


1

According to Theorem 22 in section 1.6 in this note by Pete L. Clark, the group $(\mathbb Z/p^n\mathbb Z)^*$ is cyclic, when $p$ is an odd prime. Hence we can answer your question affirmatively: Just take the algebraic closure $K$ of $\mathbb Q,$ then $K$ contains all solutions of the equation $$x^{p^n-p^{n-1}}-1=0,$$ which is a cyclic group of the same ...


1

By definition, all finite subsets of $\Omega$ are in $B$. Property 1 : $\Omega^C = \emptyset \in B$, so by definition $\Omega \in B$ since $\emptyset$ is finite. Property 2 : Note that $C = C^{C^{C}} $, so if $C \in B$ is finite then by defintion $C^C \in B$. On the other hand, if $C \in B$ is infinite, then by defintion $C^C$ must be finite for otherwise ...


0

This is not something I understand well, but I believe you can construct examples as follows. Let $E$ be an elliptic curve, and let $E'$ be another elliptic curve that is isogenous to $E$, but not isomorphic to $E$. Then the isogeny should provide an embedding $k(E') \to k(E)$, and its dual should provide an embedding $k(E) \to k(E')$.


4

A start: Let the two fields be the complex numbers $\mathbb{C}$ and the field $\mathbb{C}(X)$ of rational functions with complex coefficients. These are non-isomorphic, since one is algebraically closed and the other isn't. Embedding in one direction is trivial. In the other direction, take a transcendence base for $\mathbb{C}$ and map $X$ to an element of ...


3

This homomorphism is not well-defined. Let $g$ be the minimal polynomial of $\alpha$. Then $g(\alpha) = 0$ but $\overline g \ne 0$.


1

The first assertion is not true. For example consider the ring $O=\mathbb{Z}[\sqrt{-3}]$ in $K=\mathbb{Q}(\sqrt{-3})$. In this case $O_K = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, and $O \subsetneq O_K$, but $Frac(O)= K$. The meaning of your second question is not so clear. Are you asking if a subring of $O_K$ is integrally closed in $K$ over $\mathbb{Z}$? The ...


13

Yes, at its most basic level, a field is a generalization of the rational numbers. In a field, you can do addition, subtraction, multiplication and division as you could in $\Bbb Q$. At a deeper level, fields have geometric significance. If you've ever studied a little geometry, then you'd know that there are at least two famous ways to approach geometry: ...


6

Certain items keep showing up in mathematics. When they show up often enough, we give them names. "Field" is an example of this. You know about the real numbers: they're things that you can add and subtract, multiply and divide. Addition is commutative. So is multiplication. And multiplication distributes over addition. It turns out that there are other ...


35

More or less, yes. You can do arithmetic already with natural numbers (in $\Bbb N$), but you have to be careful with some operations: The equation $x+a=b$ does not necessarily have a solution. So one would prefer working in a ring, like the integers ($\Bbb Z$). The preceding equation has always one solution, $x=b-a$. But then, the equation $ax=b$ does not ...


2

Have you discovered yet that finite fields also satisfy the field axioms? Then the arithmetic is rather uncommon. For example $F_p$ is the field of integers modulo $p$, a prime. In $F_7$ we have things like $5\times 4=-1$. But it still has many nice properties. However, there is a theorem, basically easy, but tedious, to prove that if you also add an order ...


0

Since a^2 \in F(a) then F(a^2) \subseteq F(a). Since a is algebraic over F of odd degree, then [F(a):F] is odd (finite) ( and so F(a) is algebraic over F). So, [F(a):F] = [F(a):F(a^2)] x [F(a^2):F]. But since f(x) = x^2 - a^2 \in F(a^2)[x] has f(a) = 0, i.e., a is a root, then [F(a):F(a^2)] must divide deg(f(x)) = 2, i.e., [F(a):F(a^2)] = 1 or 2. But ...


2

Here's the plainest way I can think of: A field is a place where you can do division and multiplication commutes.


1

The isomorphism $F(a) \rightarrow F[x]/(m(x))$ just maps $a$ to $x$ (and $F$ to itself). It's actually easier to show the map the other way is an isomorphism. The map $F[x] \rightarrow F(a)$ mapping $x$ to $a$ is well-defined, and surjective because $F(a) = F[a]$. Its kernel consists of polynomials $p(x) \in F[x]$ such that $p(a)=0$. These are exactly the ...


3

We first check that if $F(t)/F$ is a simple extension and $x \in F(t) \backslash F$ then $F(t)/F(x)$ is algebraic. Indeed, write $x = P(t)/Q(t)$ and get an algebraic equation of $t$ with coefficients in $F[x]$. It follows that the whole extension $F(t)/F(x)$ is algebraic. Let now $x$, $y$ in $F(t)$. If $x$ is not in $F$ then by the above $F(t)/F(x)$ is ...


1

Let's use your idea about the degrees in a tower of extensions. We'll prove the following thing: If $a$ is a rational number such that the polynomial $X^5 - a$ is reducible over $\mathbb{Q}$ then $a$ is the fifth power of a rational number. Now, we have the following decomposition over $\mathbb{C}$ $$X^5 - a= \prod_{\eta^5 = 1} ( X- \eta \cdot a^{1/5})$$ ...


3

As has been mentioned, the result is a direct consequence of the fact that $[\mathbf{Q}(2^{1/n}):\mathbf{Q}]$ = n. This is because we then have, for every $n$, the inequality $$n = [\mathbf{Q}(2^{1/n}):\mathbf{Q}] \leq [K:\mathbf{Q}].$$ Since $[K:\mathbf{Q}]$ is larger than every natural number, it must be infinite. The trickiest point is why $2^{1/n}$ has ...


2

I’d like to expand on @tomasz”s answer in several ways. There are really three concepts here: (1) an algebraically closed field; (2) an algebraic closure of a field; and (3) the algebraic closure of one field in another. I’m going to sweep some delicate points under the rug by choosing a very handy definition of an algebraically closed field: Def. A field ...


1

Take a non zero polynomial $\;f(x)=a_0+a_1x+\ldots+a_nx^n\in L[x]\;$ , then $\;F(a_0,...,a_n)/F\;$ is algebraic (why?) . But since clearly $\;f(x)\in K[x]\;$ and $\;K\;$ is alg. closed, there exists $\;k\in K\;\;s.t.\;\;f(k)=0\;$, so $\;F(a_0,...,a_n,k)/F\;$ algebraic (why?), whick means $\;k\;$ is algebraic over $\;F\;$ and thus $\;k\in L\;$ and we're ...


2

I think the problem you have is the definition of algebraic closure of a field within an extension. Most likely the definition is as follows: the algebraic closure of $F$ in $L\supseteq F$ is the set of roots of polynomials with coefficients in $F$. The point of the exercise, it seems to me, is that this "algebraic closure in $L$" operation is indeed a ...


0

A simple extension has transcendence degree at most $1$, but $F(x,y)/F$ has transcendence degree $2$. Hope that helps,


1

It's not the best approach, but we can do this with Eisenstein's criterion: If $s=t^n$, then $X^n-s$ is irreducible as a polynomial with coefficients in $F[s]$, because $s$ is prime, the polynomial's non-leading terms are all divisible by $s$, and the constant term is not divisible by $s^2$. By Gauss's lemma, it is irreducible over $F(s)$.



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