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 ...
5
votes
1answer
223 views
Does $\varphi(1)=1$ if $\varphi$ is a field homomorphism?
Is it by definition that $\varphi(1)=1$ if $\varphi$ is a field homomorphism ?
My field theory lecture said that yes, but now $\varphi\equiv0$ is not a field homomorphism...
What is the 'standard' ...
4
votes
3answers
321 views
Equivalence of Archimedian Fields Properties
I'm trying to prove that an Archimedian field is a subfield of the real numbers, my plan is to use the fact that the rationals are dense within the field and their Dedekind completion is the real ...
3
votes
2answers
233 views
A question about a weak form of Hilbert's Nullstellensatz
Corollary 5.24 on page 67 in Atiyah-Macdonald reads as follows:
Let $k$ be a field and $B$ a finitely generated $k$-algebra. If $B$ is a field then it is a finite algebraic extension of $k$.
We know ...
1
vote
2answers
87 views
Roots of polynomials over finite fields
I've been trying to find the decomposition of $x^2-2$ to irreducible polynomials over $\mathbf{F}_5$ and $\mathbf{F}_7$.
I know that for some $a$ in $\mathbf{F}_5$ (for example), $x-a$ divides $x^2-2$ ...
1
vote
2answers
37 views
Dimension Recovery of $S \subset P_n(F)$
How is the subset of $P_n(F)$ consisting of all polynomials $f$ such that
$f(1) = 0$ a subspace of $P_n(F)$? What is the dimension of this subset?
1
vote
1answer
265 views
Cyclotomic polynomial over a finite prime field [duplicate]
Possible Duplicate:
Irreducible factors of $X^p-1$ in $(\mathbb{Z}/q \mathbb{Z})\[X\]$
Let $p$ be a prime number.
Let $l$ be an odd prime number such that $l \neq p$.
Let $X^l - 1 \in ...
1
vote
1answer
49 views
Why $g(x^{p})=(g(x))^{p}$ in the reduction mod $p$?
In one of the proof in the book "Abstract Algebra'' by Dummit and
Foote (Theorem 41, pg. 554) we have a monic polynomial $g(x)\in\mathbb{Z}[x]$,
and the book claims that $g(x^{p})=(g(x))^{p}\mod p$
...
0
votes
2answers
155 views
Necessary and Sufficient Condition for a sub-field
Is there any necessary and sufficient condition to determine whether a subset $H$ of a given field $K$ is a subfield?
In some paper I have found something like that: $H$ is a field if for all $a, ...
18
votes
2answers
362 views
What is the coproduct of fields, when it exists?
This is a slightly more advanced version of another question here.
Let $\textbf{CRing}$ be the category of commutative rings with unit. Let $\textbf{Dom}$ be the category of integral domains – by ...
16
votes
1answer
282 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 $\mathbb{Q}$?
...
12
votes
4answers
624 views
$\mathbb R^3$ is not a field
I'm trying to prove that $\mathbb R^3$ is not a field with component-wise multiplication and sum defined. I think it's weird, because every properties of a field are inherit from $\mathbb R$.
Anyone ...
8
votes
2answers
112 views
Dimension of an algebraic closure as a vector space over its base field.
Let $k$ be an infinite field and $\bar{k}$ its algebraic closure. The Artin-Schreier Theorem tells us (among other things) that $[\bar{k}:k]=1,2,\infty$. There's a natural interpretation of ...
7
votes
1answer
163 views
If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism.
I have to show the following:
Let $K$ be a field such that $\mbox{char } K \neq 2$ and each element of $K$ is a square (i.e. $K^2=K$) and let $V$ be a finite-dimensional vector spaces over $K$. ...
7
votes
2answers
286 views
Can two different roots of an irreducible polynomial generate the same extension?
Let $K$ be a field and $f(x)$ be an irreducible polynomial over $K$. Suppose, $f(x)$ has degree at least $2$. Is it possible that if $a,b$ are two roots of $f(x)$ with $a\neq b$, then $K(a)=K(b)$. ...
13
votes
3answers
239 views
Is the Pythagorean closure of $\mathbb Q$ equal to the field of constructible numbers?
A Pythagorean field is one in which every sum of two squares is again a square. $\mathbb Q$ is not Pythagorean, which is easy to see. I have read a theorem online which says that every field has a ...
9
votes
3answers
323 views
Reducibility of $P(X^2)$
This question is inspired by a comment discussion in If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism..
Let $k$ be a field of ...
8
votes
5answers
794 views
Can you construct a field with 6 elements? [duplicate]
Possible Duplicate:
Is there anything like GF(6)?
Could someone tell me if you can build a field with 6 elements.
7
votes
2answers
216 views
Elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$.
In an answer to another question I used the fact that $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$ if and only if $m$ divides $n$ (here $\zeta_n$ stands for a primitive $n$th root of unity, ...
5
votes
1answer
494 views
Problem in Jacobson's Basic Algebra (Vol. I)
It is section $4.4$, exercise number $5$. It says the following: Let $F$ be a field of characteristic $p$. Show that $f(x)=x^p-x-a$ has no multiple roots and that $f(x)$ is irreducible in $F[x]$ if ...
4
votes
2answers
142 views
When is irreducibility of a polynomial over a field equivalent to not having any roots in it?
Apart from of course, the simple cases when a polynomial $f \in K[x]$ is of degree less than or equal to three.
One direction is clear: If a polynomial is irreducible in $K$, it can have no roots in ...
2
votes
3answers
626 views
How many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$?
I know that for every $1<n\in\mathbb{N}$ there exist $p(x)\in\mathbb{F}_p$ s.t $deg(p(x))=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.
I am intrested in counting how many such $p(x)$ exist ...
13
votes
4answers
494 views
How to prove that the sum and product of two algebraic numbers is algebraic?
Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + ...
7
votes
2answers
121 views
Explicit Galois theory computation in cyclotomic field
Let $p$ be an odd prime. Let $\zeta=e^{\frac{\pi i}{4 p}}$, thus $\zeta$ is a primitive $8p$-th root of unity. There is a unique $\mathbb Q$-automorphism $\tau$ of the number field ${\mathbb ...
7
votes
5answers
550 views
Infinite Degree Algebraic Field Extensions
In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; | \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example ...
6
votes
2answers
744 views
Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?
I've been struggling to prove this fact over the past day or so.
Suppose $f(x)\in F[X]$ is irreducible over a field $F$ with $\deg(f)=n$, and let $L$ be the splitting field of $f(x)$ over $F$ with ...
6
votes
1answer
427 views
Why is it called a 'ring', why is it called a 'field'?
The definitions of 'ring' and 'field' are pretty straightforward. For a ring (e.g. integers):
addition is commutative $( 1 + 2 = 2 + 1 )$
addition and multiplication are associative $(2 +(2+2)) = ...
5
votes
1answer
290 views
Degree of field extension
If $p$ is odd prime and $c=\cos(\frac{2\pi}{p})$, $s=\sin(\frac{2\pi}{p})$ then for which values of $p$ does $\mathbb{Q}(s,c)=\mathbb{Q}(c)$?
5
votes
3answers
367 views
Algebraic Closure of Puiseux Series
Using the construction $R_N = K[t^\frac1N]$ $L_N = Quot(R_N)$ and $P = \bigcup_{N\in \mathbb{N}} L_N$ one automatically gets that the puiseux series are a field. Nevertheless they are also an ...
4
votes
1answer
166 views
Irreducible cyclotomic polynomial
I want to know if there is a way to decide if a cyclotomic polynomial is irreducible over a field $\mathbb{F}_q$?
4
votes
1answer
259 views
Is every finite separable extension of a strictly henselian DVR totally ramified?
Let $R$ be a discrete valuation ring with field of fraction $K$ and residue field $k$ and let $K'$ be a finite and separable extension of $K$.
If $R$ is henselian ("Hensel's lemma holds", e.g. if ...
3
votes
2answers
310 views
No extension to complex numbers?
Complex numbers are 2D. It is a commonly sited result that there is no 3D or 4D analogue of the complex numbers.
I just want to be clear on exactly what this result says:
It is impossible to ...
3
votes
1answer
293 views
Linearly disjoint field extensions
Recall that if $k$ is a field, some field extensions $K_1/k$,..., $K_n/k$ are called linearly disjoint if the tensor product $K_1\otimes_k\cdots \otimes_k K_n$ is a field.
Let $\zeta_5$ be a pritive ...
9
votes
4answers
338 views
Proving $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}\iff a^{2}-b$ is a square
This is an exercise for the book Abstract Algebra by Dummit and Foote
(pg. 530):
Let $F$ be a field of characteristic $\neq2$ . Let $a,b\in F$ with $b$
not a square in $F$. Prove ...
9
votes
2answers
340 views
Why is a variety over a non-alg. closed field a hypersurface?
Exercise $3$ on page $8$ of Kunz's Introduction to Commutative Algebra and Algebraic Geometry is as follows:
If the field $K$ is not algebraically closed, then any $K$-variety $V \subset A^n(K)$ can ...
8
votes
2answers
438 views
How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root?
I was trying to review some analysis, and came across problem 3 from page 78 of Walter Rudin's Principles of Mathematical Analysis. As part of the problem, I wanted to try to write ...
8
votes
1answer
180 views
Can we always find a primitive element that is a square?
Let $L/\mathbb Q$ be a galois extension. The Primitive element theorem says, that there is an element $\alpha \in L$, so that $L=\mathbb Q(\alpha)$.
Can I always find an element $\beta \in L$, so ...
8
votes
1answer
572 views
Constructive proof of the existence of an algebraic closure
It is well-known that, assuming the axiom of the choice (in the form of Zorn's lemma), one can prove that any field $F$ has an algebraic closure. One proof roughly goes as follows: consider the ...
6
votes
1answer
259 views
Why $E$ is the algebraic closure of $K$?
Let $E/K$ be a separable, algebraic extension such that every noncostant polynomial in $K[x]$ has a root in $E$, then $E$ is an algebraic closure of $K$. Could you help me to solve this exercise? ...
5
votes
6answers
137 views
Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$.
Reduction into linear factors $\mathbb{Z}_{17}[x]$:
This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so
$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv ...
5
votes
3answers
79 views
Lifting isomorphisms of fields to automorphisms of polynomial rings
Let $L$ be a field and $\alpha, \beta$ algebraic over $L$ such that $L(\alpha)\cong L(\beta)$. If $q(t)$ and $p(t)$ are the minimum polynomials of $\alpha$ and $\beta$, respectively, does it follow ...
5
votes
1answer
638 views
Finding a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$
This is exercise A4 in chapter 29 from Pinter's A Book of Abstract Algebra. It is not homework but hints/roadmap would be preferred to a full solution for now.
First some context
The book works out ...
5
votes
1answer
71 views
Verifying properties of a discrete valuation.
EDIT: Don't think about this. The problem statement is flawed. (See comments)
Let $K$ be a field, and let $K(T)$ be the quotient field of polynomials over $K$.
Then I define $v(f/g) = \deg(f) - ...
5
votes
2answers
238 views
A slick proof that a field that is finitely generated as a ring is finite
It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,...,x_n]\to k$ for some $n\in \mathbb{N}$, ...
5
votes
1answer
141 views
For which $n\in\mathbf{N}$ do we have $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$?
Put $z_{n} = e^{2\pi i /n}$. I am searching for $n \in \mathbf{N}$ so that $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$.
I know that : $z_{5} = \cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5}) $ and ...
5
votes
1answer
445 views
Least upper bound property iff convergence of Cauchy sequences
From http://en.wikipedia.org/wiki/Non-Archimedean_ordered_field:
The field of rational functions over $\mathbb{R}$ can be used to
construct an ordered field which is complete (in the sense of
...
4
votes
4answers
217 views
Show $\sqrt{2 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{2 + \sqrt{2}})$
I'm in the middle of a proof where I'd like to show that $\sqrt{2 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{2 + \sqrt{2}})$
The only way I can think of involves finding an explicit set representation for ...
4
votes
2answers
215 views
A question about a proof of a weak form of Hilbert's Nullstellensatz
I'm trying to prove the following (corollary 5.24 page 67 in Atiyah-Macdonald):
Let $k$ be a field and let $B$ be a field that is a finitely generated $k$-algebra, i.e. there is a ring homomorphism ...
4
votes
2answers
177 views
When a field extension $E\subset F$ has degree $n$, can I find the degree of the extension $E(x)\subset F(x)?$
This is not a problem I've found stated anywhere, so I'm not sure how much generality I should assume. I will try to ask my question in such a way that answers on different levels of generality could ...
4
votes
0answers
73 views
Can we construct a $\mathbb Q$-basis for the Pythagorean closure of $\mathbb Q?$
This is a follow-up question to this one. I asked it there first but moved it here following the advice from Cam McLeman.
I tried to prove that $(\mathbb P:\mathbb Q)=\aleph_0$ and I think I ...
4
votes
1answer
424 views
finding the multiplicative inverse in a field
Let $L/K$ be a field extension. Let $a\in L$ and $K[a]=\{p(a)\;|\; p\in K[x]\}$; then $K[a]$ is clearly an integral domain. I want to show that when $a$ is algebraic over $K$, then $K[a]$ is a field.
...
