All Questions
Tagged with finite-rings finite-fields
21 questions
4
votes
2
answers
436
views
Do there exist finite commutative rings with identity that are not Bézout rings?
A similar question has been asked before: Example of finite ring which is not a Bézout ring, but has not been answered.
There also seems to be a dearth of resources online regarding this ...
3
votes
0
answers
102
views
Show that $\Bbb Z_p[i]$ is isomorphic to $\Bbb Z_p[x]/\langle x^2+1\rangle$.
Let $p$ be a prime number. Show that $\Bbb Z_p[i]$ is isomorphic to $\Bbb Z_p[x]/\langle x^2+1\rangle$.
My attempt:
Define
$$\phi : \Bbb Z_p[x] \to \Bbb Z_p[i]$$
by $\phi\big(f(x)\big)=f(i)$. ...
0
votes
1
answer
3k
views
Finite integral domain is a field: proof condition
Theorem: a finite integral domain is a field
proof:
Let D be a finite integral domain with unity 1. Let a be any non-zero element of D.
If a=1, a is its own inverse and the proof concludes.
Suppose,...
0
votes
0
answers
267
views
What is the difference between $Z/(4)$ and the field $F4$? [duplicate]
I assumed that if we quotient $Z$ by the ideal generated by 4, it will be the same as the field $F4$. It turns out not to be the case because $Z/(4)$ is not a field.
In general, when is $Z/(n) = Fn$?
0
votes
1
answer
491
views
How to describe a ring after adjoining an element to it?
I want to know how to go about describing a ring after adjoining an element that satisfies a certain relation.
As an example, I'm considering the ring obtained from Z3 by adjoining an element a ...
0
votes
2
answers
922
views
Definition of a Quotient Ring (need clarification).
My understanding is that we can view the definition of a Quotient ring $R/I$ as a set of cosets.
For example, the ring $Z/(6)$ which I believe is $Z6$, can be viewed like this:
$(6) + 0 = \{...,-12,...
0
votes
2
answers
337
views
Factoring polynomials in $\mathbb Q[x]$ and $\mathbb Z[i]$
I'm practicing reducing polynomials in different rings, but I'm stuck on a few. I would appreciate any help.
1) I want to factor $7+i$ in $\mathbb Z[i]$. The norm is $50$, so we know it's reducible....
1
vote
0
answers
180
views
Prove that $\mathbb{Z}[i]$ is a euclidean domain (intuition behind the geometric proof)
I am viewing the proof that $\mathbb{Z}[i]$ is a Euclidean domain, but I'm having a very hard time imagining it geometrically. Or in other words, showing that the size function exists by viewing the ...
1
vote
2
answers
316
views
Extension of a finite field to a finite non commutative ring
Can a finite field be extended to non-commutative finite rings so that not all elements of the field commutes with the elements of the ring?
I have been trying this taking the examples of matrices.
0
votes
2
answers
81
views
On analogy between $\Bbb Z$ and $\Bbb F_q[x]$
There are objects and operations analogous between $\Bbb Z$ and $\Bbb F_q[x]$. For example primes in $\Bbb Z$ and irreducibles in $\Bbb F_q[x]$ are analogous and so is multiplication operation.
...
0
votes
2
answers
2k
views
Finite integral domain
I encountered a problem:
Every finite integral domain is isomorphic to $ \mathbb{ Z }_{p} $.
I know that finite integral domain is isomorphic to a field, but I have no idea on how to construct a ...
4
votes
3
answers
329
views
How many elements does this ring have?
I know that the following ring is not a field because the defining polynomial is reducible into two polynomials that are irreducible:
$$\mathbb Z_2[X]/(x^5+x+1)$$
where
$$x^5 + x + 1 = (x^2 + x + 1) (...
2
votes
1
answer
124
views
Find a ring homomorphism $\tau: \mathbb{F} \rightarrow \mathbb{F}$
Just working on some exam prep questions, and I'm a bit stuck on this one.
Let $ \mathbb{F} = \{ a + bX + cX^2 | a,b,c \in \mathbb{F}_2 = \{0,1\} \} $ be a ring with the operations:
Addition, ...
2
votes
0
answers
101
views
Finite Ring with unity and no zero divisors is field [duplicate]
I would like to know if someone can help me with this.
"Show that a finite ring $R$ with unity $1\neq 0$ and no divisors of 0 is a field."
The original exercise asked me to show that it was a ...
1
vote
2
answers
506
views
Why is $1 + 1 = 0$ in $\{0, 1\}$ (binary field) and not 1 or 2?
Stuck on the simplest case in my foray into fields...
I know there is a really similar question out there, but I can't find any contradiction with the field axioms if 1 + 1 = 1 instead of 0.
Can ...
3
votes
2
answers
1k
views
Can anyone explain the difference between a ring, a group, and a field in a way so that your average 15 year old can understand? [duplicate]
I can not understand key difference between them, because all on the web uses mathematical notations...Can anyone explain in easy way?...please.Thanks in advance.
2
votes
2
answers
2k
views
$\sum_{x \in \mathbb Z_9}w^{x^2-cx}=?$,where $c \in \mathbb Z_9$ and $w=e^{2\pi i/9}$
$$\sum_{x \in \mathbb Z_9}w^{x^2-cx}=?$$
where $c \in \mathbb Z_9$, $w=e^{2\pi i/9}$ and $\mathbb Z_9$ is the ring of integers modulo 9.
0
votes
1
answer
64
views
Is $\sum_{c \notin \mathbb Z_q}\psi(c)-\sum_{c \notin \mathbb Z_q}\psi(c^n)$ an integer?
Let $n$ be a positive integer. Let $p$ be an odd prime and $q=p^k$. Let $c \in \mathbb Z_q$. Consider the additive character $\psi:\mathbb Z_q \rightarrow \mathbb C^{\times}$ that is defined as $\psi(...
18
votes
3
answers
17k
views
How to show that a finite commutative ring without zero divisors is a field?
$R$ is finite commutative ring without zero divisors which has at least two elements. How to show that $R$ is field?
I'm just starting with abstract algebra and I'd really appreciate if someone could ...
1
vote
1
answer
2k
views
How many orthogonal matrices are there over a given finite ring or field?
I want to know how many $2\times 2$ orthogonal matrices exist over the ring $\mathbb{Z}_n$ or the field $\mathbf{F}_p$. And how many $2\times 1$ orthogonal vectors exist over the ring $\mathbb{Z}_n$ ...
33
votes
1
answer
2k
views
Universal binary operation and finite fields (ring)
Take Boolean Algebra for instance, the underlying finite field/ring $0, 1, \{AND, OR\}$ is equivalent to $ 0, 1, \{NAND\} $ or $ 0, 1, \{ NOR \}$ where NAND and NOR are considered as universal gates. ...