# Tagged Questions

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....

545 views

### An exercise with Zariski topology

I read this exercise: Prove that the set $S = \{ (n, 2^n, 3^n ) \mid n \in \mathbb{N} \}$ is dense in $\mathbb{C}^3$ with Zariski topology. I have seriously thought about it, but I do not manage to ...
693 views

### Irreducibility of $x^{n}+x+1$

Motivated by this problem, and KCd's comment on my answer, I am left with the following question: Question: Suppose that $n\not \equiv 2\pmod{3}$. Is $$x^n+x+1$$ irreducible over $\mathbb{Q}$? ...
706 views

### Why isn't the perfect closure separable?

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is ...
2k views

### A finite field cannot be an ordered field.

I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to $\mathbb R$. Since all ordered finite fields would have supremum property that must mean none exist. ...
486 views

### Is every rigid field perfect?

A field is rigid iff its automorphism group is trivial. A field $F$ is perfect iff all irreducibles in $F[x]$ are separable. Is every rigid field perfect?
709 views

### Finding the degree of a field extension over the rationals

Let $\alpha = \sqrt{\sqrt{2}+\root 4 \of 2}$, $\beta =\sqrt{\sqrt{2}- \root 4 \of 2}$, $\gamma = \sqrt{-\sqrt{2}+i\root 4 \of 2}$ and $\delta = \overline{\gamma}=\sqrt{-\sqrt{2}-i\root 4 \of 2}$. Let ...
954 views

### Galois group of a reducible polynomial over $\mathbb {Q}$

Let $f \in\mathbb {Q}[X]$ be reducible - for the sake of simplicity, $f = gh$ with $g,h \in\mathbb {Q}[X]$ irreducible. Let L be the splitting field of f. Does $Gal(f) \simeq Gal(g) \times Gal(h)$ ...
567 views

3k views

### Proving the “freshman's dream”

$(x+y)^p = x^p + y^p$ holds in any field of characteristic $p$. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems. What is the ...
462 views

### Why does $K \leadsto K(X)$ preserve the degree of field extensions?

The following is a problem in an algebra textbook, probably a well-known fact, but I just don't know how to Google it. Let $K/k$ be a finite field extension. Then $K(X)/k(X)$ is also finite with ...
1k views

### How to transform a general higher degree five or higher equation to normal form?

This is a continuation of How to solve fifth-degree equations by elliptic functions? How to transform a general higher degree five or higher equation to normal form? For xeample, a quintic equation ...
377 views

### A sentence false in a field of characteristic $0$ but true in all fields of positive characteristic?

Consider the language $L=\{+,\cdot, 0, 1\}$ of rings. It is easy to show using compactness that if $\sigma$ is a sentence that holds in all fields of characteristic $0$, there is some $N\in \mathbb N$ ...
2k views

### Basis of primitive nth Roots in a Cyclotomic Extension?

While reading one of Keith Conrad's great blurbs, Linear Independence of Characters, there is a footnote at the bottom of page 2 saying In general, the primitive $n$th roots of unity in the $n$th ...
419 views

3k views

### Every algebraic extension of a perfect field is separable and perfect

I am trying to prove this statement in the characteristic $p>0$ case. Every algebraic extension of a perfect field is separable and perfect. This is stated as a corollary of Proposition V.6....
758 views

252 views

522 views

### Topological Quantum Field theories

I was wondering about the following on TQFTs. It is said that TQFTs have vanishing Hamiltonians $\hat{\mathcal{H}}$. Firstly, I would like to ask: Why is this so? Secondly, consider the ...
327 views

### Fields whose algebraic closure cannot be constructed without the axiom of choice

One can show that the statement that every field has an algebraic closure requires the axiom of choice. However, for almost all "everyday" fields, it seems that one can actually produce an algebraic ...
804 views

### Example of a category without product

A question was raised in our class about the non-existence of product in a category. The two examples that came up in the discussion was the category of smooth manifolds with boundary and the category ...
$R$ is a domain with characteristic $p$ ($p$ is prime).There is a homomorphism $f : R \to R$, $f(a)=a^p$. $f$ is called the Frobenius homomorphism. And I have known this. When $R$ which is mentioned ...