# Questions tagged [field-theory]

Use this tag for questions about fields and field theory in abstract algebra. A field is, roughly speaking, an algebraic structure in which addition, subtraction, multiplication, and division of elements are well-defined. Please use (galois-theory) instead for questions specifically about that topic. Use (classical-mechanics), (electromagnetism), (general gravity) or (quantum-field-theory) for physical fields.

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### In ZFC there exists a perfect field of given positive characteristic and given infinite cardinality

Fix a prime number $p$. In ZFC does there exist a perfect field of characteristic $p$ of any infinite cardinality? I know some constructions of fields of characteristic $0$ of arbitrary cardinality ...
0answers
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### Why $X$ is a primitive r-th root of unity in lemma 4.7 Prime is in P? [duplicate]

Let $Q_r(X)$ be the $r^{th}$ cyclotomic polynomial over $F_p$. Polynomial $Q_r(X)$ divides $X^r-1$ and factors into irreducible factors of degree $o_r(p)$. Let $h(X)$ be one such irreducible factor. ...
0answers
20 views

### Question about finite extensions and splitting fields

Apparently any finite extension of a field with characteristic 0 is of the form $F(a)$ for some algebraic $a$. But wouldn't this mean that any two zeros of a polynomial (e. g. over $\mathbb{Q}$) would ...
1answer
56 views

### Show that $\mathbb{Q}(\sin\theta)$ is a field

QUESTION: Show that $F_{\theta}=\mathbb{Q}(\sin\theta); \theta\in\mathbb{R}$ is a field. Moreover, Show that $E_{\theta}=\mathbb{Q}(\sin\frac{\theta}{3}); \theta\in\mathbb{R}$ is a field extension ...
1answer
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### Residue field and value group of an algebraically closed valued field [closed]

Why is the residue field of an algebraically closed valued field algebraically closed and the value group is divisible?
1answer
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### Galois Group of $\mathbb{Q}(\sqrt{\sqrt{3}-1})$

I am reviewing Galois Theory questions for an upcoming prelim, and I came across this question on a past prelim "Find the Galois Group $\operatorname{Gal}(\mathbb{Q}(\sqrt{\sqrt{3}-1})$" The minimal ...
2answers
46 views

### Elements of $E^{\times},\cdot$ of the quotient ring $E:= \frac{\mathbb{Z}_3[X]}{\langle x^2 + x + 2\rangle}$

Consider the field $E:= \frac{\mathbb{Z}_3[X]}{\langle x^2 + x + 2\rangle}$. If I'm right the elements of the quotient ring can be found as: $$a_0 + a_1x + \langle x^2 + x + 2\rangle.$$ So we got the ...
5answers
220 views

### Show that $\sqrt{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt{2})$

Setting $\alpha = \sqrt{2}$ and $a+b\alpha+c\alpha^2=\sqrt{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ ...
1answer
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### Proof Explanation *Field and Galois Theory*

The following is with regards to Lemma 3.18 from Field and Galois Theory by Patrick Morandi. Let $\sigma : F \to F'$ be a field isomorphism, $K$ be a field extension of F, and $K'$ be a field ...
1answer
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### Galois group of order $n!$

Suppose $f\in\mathbb{Q}[x]$ has degree $n$, and let $K$ be the splitting field of $f$ over $\mathbb{Q}$. Suppose the Galois group of $K/\mathbb{Q}$ is $S_n$ with $n\geq 3$. It is not hard to show that ...
1answer
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### Prove that $[\mathbb{Q}(\sqrt{3},i):\mathbb{Q}=4$. Find a number $a$, such that $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{3},i)$

The first part follows from the fact that for any algebraic elements $a_i$ the following holds: $$F(a_1,...,a_n)=(F(a_1,...,a_k))(a_{k+1},...,a_n)$$ where $F$ is the field generated by the $a_i$. ...
3answers
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### Degree of the splitting field over finite field

Let the K be a splitting field of $f(x) =x^3 + 5x+ 5$ over $Z_3$ What is the $[K;Z_3]$ ? The solution in my book) Let the $\alpha$ be a solution of the $f(x$) Since $K = Z_3 (\alpha$) ...
1answer
28 views

### Let $K/F$ be a field extension. If $\alpha \in F(\alpha^m)$, $m > 1$, then $\alpha$ is algebraic in $F$.

Let $K/F$ be a field extension. If $\alpha \in F(\alpha^m)$, $m > 1$, then $\alpha$ is algebraic in $F$. A proof provided in the book is as follows: Proof: Since $\alpha \in F(\alpha^m)$, there ...
0answers
48 views

### $X^{p^n} - a$ is irreducible if $a \in K$ has no $p$-th root? [duplicate]

Exercise from Lang. I am trying to show that if $char K = p$ and $a \in K$ has no $p^{th}$ root (implying that K is an infinite field) then $X^{p^n} - a$ is irreducible for all positive integer $n.$ ...
1answer
34 views

### Finding elements that are algebraic over a given field

Find all $k \in \mathbb{N}$ such that there exist elements in the field $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2)}$ that are algebraic of order $k$ over $\mathbb{Q}$. For each such $k$ find an ...
3answers
58 views

### If every non-zero ring homomorphism $\phi: K\rightarrow S$ is injective, $K$ is field

If every ring homomorphism $\phi: K\rightarrow S$ is injective, $K$ is field. PS: $K$ is a commutative ring with unity, and my definition of homomorphism includes $\phi(1_{K})=1_{S}$. My idea is to ...
2answers
34 views

### In a finite field there exists an irreducible polynomial of degree at least $n$ for $n \in \mathbb{N}$

This question already has an answer here, but I'm looking for a solution that doesn't use field extensions. It's relatively easy to find a polynomial without zeros for every $n$ but as far as I know ...
0answers
61 views

### Normal extension $\Longleftrightarrow$ splitting field extension, intuitive or counterintuitive? [on hold]

$\textbf{Question}$: Would you say that normal extension $\Longleftrightarrow$ splitting field extension is intuitive or counterintuitive? A splitting field extension $L:K$ is exactly large enough to ...
1answer
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### When is the Galois group of a quintic not $S_5$ if we use this particular method?

I think that I am misunderstanding something fundamental about the technique used to decide if higher order polynomials are solvable by radicals using Galois theory. If we have a cubic it's not to bad ...
2answers
134 views

### Is every $1$-dimensional vector space a field?

We say that every field $F$ "is" a $1$-D vector space over itself. By this we mean that if we consider the elements of $F$ as both vectors and scalars, then we get a vector space by using the addition ...
0answers
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### Why isn't my choice of element fixed by the automorphism of this subgroup of $Gal(\Bbb Q(w_{20})/Q)$?

Given the twentieth root of unity a splitting for it's minimal polynomial is clearly $\Bbb Q(w_{20})$ where $w_{20}$ is the twentieth primitive root of unity. As the isomorphisms only map to roots of ...
1answer
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### Proving the following automorphism of fields is an additive homomorphism

I have the following question, if $\varphi: \mathbb{F}_{p^2} \longrightarrow \mathbb{F}_{p^2}$ defined as $\varphi(x) \mapsto x^p$ is an automorphism, then how do I show it is a homomorphism in the ...
2answers
47 views

### If $a$ is a sum of squares then there exists an element whose norm is $-1$.

This is part of a question(2.9.3) from Patrick Morandi's "Field and Galois Theory" I am just plain stuck on this fact: $F(\sqrt{a})/F$ is a field extension where $a\in F-F^2$ and $F$ does not ...
2answers
94 views

### The number of roots of a polynomial $p(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{11^2}$.

I am trying to count the number of roots of a polynomial $p(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{11^2}$. Of course, I can plug every element in $\mathbb{F}_{11^2}$ into $x$ in $p(x)$. But is there any ...
1answer
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0answers
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### Show that a homomorphism from $SL_2(\mathbb{Z})$ to $SL_2(\mathbb{F}_p)$ is surjective

The homomorphism sends the entries of matrices in $SL_2(\mathbb{Z})$ to their congruence classes mod $p$. After a lot of work I could prove that a matrix \begin{pmatrix}a & b\\\ c & d\end{...
2answers
49 views