Are the units of the quotient ring $\mathbb{F}_2[x]/\langle x^k+1 \rangle$ known in general, where $\mathbb{F}_2$ is the finite field with two elements? I'm specifically interested in the case where $k$ is divisible by two, such as say $k=8$, or $k=12$.  If so, can they be constructed easily? I'd be grateful if you could provide a few examples.
120863
Are the units of the quotient ring $\mathbb{F}_2[x]/\langle x^k+1 \rangle$ known in general, where $\mathbb{F}_2$ is the finite field with two elements? I'm specifically interested in the case where $k$ is divisible by two, such as say $k=8$, or $k=12$.  If so, can they be constructed easily? I'd be grateful if you could provide a few examples.
3350
I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.  Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.  Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.   I have not been able to get the converse part and would like to see the proof for the converse part.
108725
maybe that's an idiot question. Given a finite field $k$ and some irreducible polynomial $f(x) \in k[x]$, then $k_f \cong k[x]/(f(x))$? I know that it's true if $k$ is the prime field and I think that the statement is not true for general finite fields, however I could not find any counter example.  Thanks in advance.
13785
I have got stuck with two generator problems:        The ideal $(zx,xy,yz)$ can't be generated by $2$ elements.    The ideal $(xz-y^2,yz-x^3,z^2-xy)$ can't be generated by $2$ elements.      Here the ring is $k[x,y,z]$.  Thanks in advance for any help.
67273
I had to calculate the number of elements of this quotient ring:   $$R = \mathbb{Z}[X]/(X^2-3, 2X+4).$$   This is what I've got by myself and by using an internet source:  Writing the ring $R = \mathbb{Z}[X]/(X^2−3, 2X+4)$ as $\mathbb{Z}[X]/I$, we note that $I $ contains $2(X^2−3)−(X−2)(2X+4) = 2$. We then note that the generator $2X + 4$ is actually superfluous since $2X + 4 = 2(X + 2)$.   Now we can write $R = \mathbb{Z}[X]/(X^2 − 3, 2) = \mathbb{Z}[x]/((X+1)^2, 2)$, because $(X+1)^2=X^2+2X+1=X^2-3+2X+4$. The internet source states now the following:  $$ R \cong (\mathbb{Z}/2\mathbb{Z})[\alpha] \quad \text{with} \quad \alpha = X+1$$ I guess that $\mathbb{Z}/2\mathbb{Z}[\alpha]$ represents the set of dual numbers of the field $\mathbb{Z}/2\mathbb{Z}$. I see that $\alpha^2=0$, but what exactly implies the isomorphism? And does this mean that the quotient ring $R$ contains four elements?  If necessary, you can take at the site I used. (It's about page 5, exercise 4.3a.)    I thank you in advance for your answers.
13573
I'm currently considering the ideal $(2,X+1)\subset\Bbb{Z}[X]$. I'm trying to figure out if this is a prime ideal, a maximal ideal, or neither. So far I have attempted to look at the quotient ring $\Bbb{Z}[X]/(2,X+1)$. I know that if this quotient ring is a field, then our ideal is maximal, since $\Bbb{Z}[X]$ is commutative. However I'm stuck on how I should be viewing this quotient ring.
15433
Is there a nice way to perform calculations within a quotient ring? for instance, an exercise I had was to show $\mathbb{Z}[X]/(X^2+1) \cong \mathbb{Z}[i]$, but this got me thinking; If I didn't know this, how could I calculate in $\mathbb{Z}[X]/(X^2+1)$ easily?  (and of course any other general quotient ring)?
68270
Motivated by :   Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of units of some commutative ring? Which finite abelian groups are the group of units of some finite commutative ring?   It  that even for fields this is not simple to answer.  I expect the general question (Which groups are the group of units of some ring?, with no finiteness hypotheses) is even harder.
52211
In a ring If product of finite elements is a unit then can we say each element as unit  For ex., if $q_1 q_2 ... q_m$ is a unit, then can we say each of $q_i$ $\forall i$ is a unit how to prove this?
106059
I've been considering the rings $R_1=\mathbb{F}_p[x]/(x^2-2)$ and $R_2=\mathbb{F}_p[x]/(x^2-3)$, where $\mathbb{F}_p=\mathbb{Z}/(p)$.   I'm trying to figure out if they're isomorphic (as rings I suppose) or not for primes $p=2,5,11$.   I don't think they are for $p=11$, since $x^2-2$ is irreducible over $\mathbb{F}_{11}$, so $R_1$ is a field. But $x^2-3$ has $x=5,6$ as solution in $\mathbb{F}_{11}$, so $R_2$ is not even a domain.  For $p=5$, both polynomials are irreducible, so both rings are fields with $25$ elements.  I know from my previous studies that any two finite fields of the same order are isomorphic, but I'm curious if there is a simpler way to show the isomorphism in this case, without resorting to that theorem.  For $p=2$, neither ring is even a domain as $x=0$ is a solution of $x^2-2$ and $x=1$ is a solution for $x^2-3$, but I'm not sure how to proceed after that. Thank you for any help.
41499
Let $K$ be a field and $R=K[X]/(X^n)$ where $n \in \mathbb{Z}_{n\geq1}$ and $(X^n)$ is the ideal generated by $X^n$. We denote $x:=X+(X^n) \in R$, any equivalence class $r$ in $R$ has a representing element of the form: $$r=a_0+a_1X+ \dots + a_{n-1}X^{n-1}.$$  I am confused about the statement in bold. How would I find such an element for $X^3 \in K[X]$ if $n=2$ for example?
2076
Where $F$,$K$ are two fields. What does the element in the composite field $FK$ look like? All the elements are generated by the elements of $F$ and $K$? (combination of the elements of $F$ and $K$) I think $FK$ is pretty close to the free product of $F$ and $K$; is $FK \subsetneq F*K$? Thank you!
15429
Can someone please give me a hint how to prove that the quotient ring $\mathbb{F}_p[x]/{\langle f\rangle}$, where $f$ is a irreducible polynomial of degree $k$ and $p$ is a prime (and $\langle f\rangle$ the ideal generated by $f$), has cardinality $p^k$ and how to determine the dimension of $\mathbb{F}_p[x]/{\langle f\rangle}$ as a vector space ?
36435
Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements  The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is congruent to 3 mod 4 it is irreducible thus prime. and hence the ideal $(q)$ is prime and therefore is maximal because $\mathbb{Z}[i]$ is a pid. Thus $\mathbb{Z}[i]/(q)$ is a field  I am not totally clear on the number of elements being $q^2$. I think perhaps the division algorithm would help but I am still not totally clear on how to use it in a case like this.
54489
In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?  Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?
40010
Let $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$ denote the field of 2 elements   (a) Is $x^4+x^2+1$ irreducible in $\mathbb{F}_2[x]$? Find a complete factorization.  (b) How many irreducible polynomials of degree 4 are there in $\mathbb{F}_2[x]$?  For the first part of the problem, I just plug in $0$ and $1$ into $x^4+x^2+1$, and know that there is no root for that polynomial, so this polynomial is irreducible.   For the second part, I just list polynomials of degree 4 and having no zero in $\mathbb{F}_2[x]$, which are $x^4 + x+1, x^4 +x^2 +1, x^4 + x^3 +1$ , and $x^4 + x^3 +x^2 + x +1$.   But I am not sure about whether these approaches are right, so could you help me to figure it out? Thank you!
63044
This should be very basic but I am having a bit of trouble finding the field of fractions for quotients of polynomial rings over a field. The specific example I am having trouble with is the following:  Let $k$ be a field with characteristic $\neq 2$ and let $f \in k[x_1, \dots , x_n]$ be a polynomial with no repeated factors. I am trying to determine the field of fractions of the ring $$ k[x_1, x_2, \ldots , x_n, z]/\langle z^2 - f\rangle. $$  The usual method I would use for a problem like this is the Chinese remainder theorem, but I don't see any way to apply that here. The other fact I know is that if $A$ is an integral domain with prime ideal $\mathfrak{p}$, then $\text{Frac}(A/\mathfrak{p}) \simeq A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$. But the problem of finding  $A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ seems no easier in this case. Can anyone point me in the right direction? I am fairly certain I know what the result is, but I have no idea how to show it properly.