New answers tagged ring-theory
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different definitions of Hopf algebras
This question has been successfully answered if you combine the other answers and the comment discussion that they contain. I will summarize:
There are various differences between the definitions. ...
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Question on proof that all subgroups of $\mathbb{Z}$ and $\mathbb{Z_n}$ are subrings (and ideals).
Let's prove that every subgroup $A$ of $\mathbb{Z}$ is an ideal. We need to show:
$0 \in A$
$\forall a \in A: -a \in A$
$\forall a,b \in A: a+b \in A$
$\forall n \in \mathbb{Z}, a \in A: na \in A$
...
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Question on proof that all subgroups of $\mathbb{Z}$ and $\mathbb{Z_n}$ are subrings (and ideals).
Let $G$ be a subgroup of $\mathbb{Z}$. Let $a,b\in G$, then $a+b,-a\in G$ because it is a subgroup. $ab=a+...+a\in G$ for the same reason. This makes $G$ a sub-rng (ring without $1$) of $\mathbb{Z}$ - ...
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Is "Zariski connected" equivalent to "ring connected"?
After a little thought, this does turn out to be equivalent to the definition of "connected" that I have seen, which is "no nontrivial idempotents."
It's known that $\mathrm{Spec}(...
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Examples of subgroups of additive group that aren't subrings
$\mathbb{Z}$ is a ring under normal addition and multiplication but the subgroups $2 \mathbb{Z}$, $3 \mathbb{Z}$, ... $p \mathbb{Z}$ are not (no unit).
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Examples of subgroups of additive group that aren't subrings
Consider the group of $2\times 2$ real matrices, if you consider the additive subgroup generated by the matrix
$$
\begin{bmatrix}
1 &1\\ 0 & 1
\end{bmatrix}
$$
The additive group it generates ...
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Order of Odd Elements in $\mathbb{Z}_{2^n}$
There is a very easy method for each $a$ where $a \equiv 1 \bmod 4$ and $a \not\equiv 1 \bmod 2^n$. Find the highest power of $2$ dividing $a - 1$, say $a - 1 = 2^kb$ where $2 \leq k \leq n-1$ and $b$ ...
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Order of Odd Elements in $\mathbb{Z}_{2^n}$
For the case $n=8$ you can just brute force it (there are only 128 odd numbers $\mod 2^8$).
You can easily compute the multiplicative order of an element in $\mathbb{Z}/2^n$, but there is no known ...
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Order of Odd Elements in $\mathbb{Z}_{2^n}$
Not really, and that's not a phenomenon that's restricted to $\Bbb Z_{2^n}$—calculating orders modulo $m$ is untidy in general. When $n>2$, every odd element in $\Bbb Z_{2^n}$ can be written ...
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projective and injective ideals of $\mathbb Z/n\mathbb Z$ as $\mathbb Z/n\mathbb Z$ module
It's known that $\mathbb Z/n\mathbb Z$ is always a quasi-Frobenius ring, and since the injective modules coincide with the projective modules, solving one problem solves the other.
Considering that $R$...
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$X$ be a connected metric space and $F$ be subring of $C(X,\Bbb{R})$ which is a field. Then every element of $F$ is constant. (T/F)
I will adopt the usual definition that a ring contains a multiplicative identity and that a subring must inherit the same identity as its super-ring.
Note that the constant function $\mathbf{1}$ lies ...
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Prove that $\sqrt{-5}$ is a prime in the ring $R=ℤ[\sqrt{-5}]$.
This is what you need to show in order to prove that $\sqrt{-5}$ is a prime in $\mathbb Z[\sqrt{-5}]$:
For any $a, b \in \mathbb Z[\sqrt{-5}]$, if $\sqrt{-5}$ divides $ab$, then $\sqrt{-5}$ divides $...
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Zero divisors in $\mathbb Q [T]/(T^4+T^2+1)$
Actually, $T^4 + T^2 + 1$ is reducible in $\mathbb Q[T]$, since
$$ T^4 + T^2 + 1 = (T^2 + T + 1)(T^2 - T + 1).$$
In the quotient ring, $\mathbb Q[T]/(T^4 + T^2 + 1)$, the elements $$a = [T^2 + T + 1]_{...
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Induced filtration on polynomial ring with coefficients in a filtered associative algebra
It depends.
If you really consider $t$ to have degree $0$ and therefore to have $k[t]$ unfiltered (or rather with trivial filtration), then the induced filtration on $A[t]$ should be the first one.
If ...
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Commutative algebra: Irreducible components of $\operatorname{Spec}(A)$
The minimal primes of $A$ correspond to the minimal primes over $I$. If $P$ is a minimal prime containing $I$, then $X(Y+1)\in P$, so either $X\in P$ or $Y+1\in P$. If $X\in P$, then $I\subseteq (X)\...
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Non-zero idempotent element in a non-nilpotent left artinian ring
Steven Gaal's Linear Analysis.. states in proposition I.7.11 that
If $A$ is a ring with identity satisfying the minimum condition for left ideals then every non-nilpotent left ideal contains an ...
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Fraction field of $\mathbb Z_p[[X]]$
$\newcommand{\Z}{\mathbb{Z}}$
Every element $f\in\mathbb{Z}_p[[x]]$ can be written as
$$
f(x) = p^e u(x) q(x)
$$
with $e \geq 0$, $u(x)\in \mathbb{Z}_p[[x]]^*$ and $q(x)\in \mathbb{Z}_p[x]$ monic.
...
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Confused on Proof that Polynomials of Degree $n$ has at most $n$ zeroes (counting multiplicity) from Gallian's Contemporary Abstract Algebra
I would have used the same proof as you suggested, but I realized that it has a little gap: We have to show that all zeros of $f(x)$ are zeros of one of its factors. This seems to be trivial; ...
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If there exists $n>1$ such that $x^n=x$ for all $x$ in a ring, then there are no nonzero nilpotent elements.
"Suppose that there is an integer $n > 1,$ such that $x^n = x$ for all elements $x$ of some ring",
it is actually means the ring has trival multiplicative/additive subgroup {0},
and all ...
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Let $p$ be a prime $\equiv 1 \pmod 4$. Prove that $\Bbb{Z}[i]/(p)$ $\cong$ $\Bbb Z_p \times\Bbb Z_p$
We can prove that $\Bbb{Z}[i]/(p) \cong \Bbb{Z}_p \times \Bbb{Z}_p$ using the Chinese Remainder Theorem and the fact that $p$ is a prime $\equiv 1 \pmod 4$.
First, note that $\Bbb{Z}[i]$ is a ...
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Let $p$ be a prime $\equiv 1 \pmod 4$. Prove that $\Bbb{Z}[i]/(p)$ $\cong$ $\Bbb Z_p \times\Bbb Z_p$
Seems like one way to go, assuming you know what the primes of $\mathbb{Z}[i]$ are, would be to apply the Chinese remainder theorem by noticing that, since $p \equiv 1 \mod 4$, there exist $(a,b) \in \...
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Let $p$ be a prime $\equiv 1 \pmod 4$. Prove that $\Bbb{Z}[i]/(p)$ $\cong$ $\Bbb Z_p \times\Bbb Z_p$
Consider $f(a+b\ i) = (a+b\ u, a-b\ u)$ where $u^2\equiv -1 \pmod p$ ($u$ exists since $p \equiv 1 \pmod 4$)
To see where this come from, observe that since $f$ is a ring homomorphism we have:
(All ...
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Find kernel of homomorphism
Very good start! In fact, this is biconditional: $f(a+bi) = 0$ if and only if $a = b$ in $\mathbb{Z}_2$. Now we fully understand $\ker(f)$!
Next, we need to fully understand $(1+i)\mathbb{Z}[i]$. An ...
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Group ring over a cyclic group
The hint in the comments is a good one that works for cyclic groups, but there is actually one that works simply for all finite groups (or, for that matter, any group with a nontrivial finite ...
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how to find units for the ring Z[1/2]
The units of $\mathbb{Z}[1/2]$ are precisely the signed powers of two and their opposites, that is: $$\mathcal{U}_{\mathbb{Z}[1/2]} = \{(-1)^n2^k:n\in\mathbb{N},k\in\mathbb{Z}\}$$
We can prove this ...
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UFD implies GCD
As the previous answers have addresses your problem(and clearly the proof you posted is, admittedly, confusing), I just wish to provide a clean and easy-to-read proof to my taste. Our goal is to show ...
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When is a quotient $\mathbb Z$-algebra a ring quotient
Note that if $f:S\to R$ is a surjective and multiplicative map between two rings $S$ and $R$ with identity, then $f(1_S)=1_R$. Indeed, if $s$ is a preimage of $1$, then
$$
1=f(s)=f(s \cdot 1)=f(s)f(1)=...
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Exercise about Ideals, Generators, principal ideals and prime ideals
I'll start with correct proofs of each of the claims (assuming I didn't make a mistake), and then comment on each section of your question.
I'm putting my answers first because that's the order in ...
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Exercise about Ideals, Generators, principal ideals and prime ideals
You seem to be using $M$, $A$, and $I$ for the same set here. Stick with $A$ as given in the original problem. Edit Now I see that $M$ stands for something different; a set of generators.
Your answer ...
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Prove this is an exact sequence
If one of $a,b$ is 0, then the proof is easy.
So assume $a,b\not=0$.
The last map in the sequence should be
$$f: ao + bo \longrightarrow o/(ao:bo)$$
$$ as + br \mapsto r$$
To see $f$ is well-defined, ...
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Accepted
Proof that gaussian integers are an integral domain by contradiction.
Although it is already too late, maybe there is a better way to see why your proof is wrong. You see, one of the reasons we want to show something is an integral domain is that cancellation property ...
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Let $f:A→B$ be a ring morphism and $φ : \operatorname{Spec}B→\operatorname{Spec}A$ its induced continuous map, then prove that $\rm im\ φ ⊆ V(\ker f)$
I came up with this solution:
Let $P\in \operatorname{Spec}(B)$ lets see that $\phi(P)\in V(\ker (f)) \:$, as $P$ is an ideal we have that $0 \in P\:$ so $f^{-1}(0)\subset f^{-1}(P) \:$ that means ...
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Let $f:A→B$ be a ring morphism and $φ : \operatorname{Spec}B→\operatorname{Spec}A$ its induced continuous map, then prove that $\rm im\ φ ⊆ V(\ker f)$
Strong hint:
This is essentially the correspondence theorem but for rings rather than groups. Namely, ideals in $A$ that contain $\ker f$ are in correspondence with ideals in $B$, under the same ...
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If $R$ is a Noetherian ring, then $R^n$ is Noetherian
If $R$ is noetherian, the polynomial ring $R[x_1,\dotsc,x_n]$ is also noetherian. Consider the ideals
$$I_k = (x_1, x_2, \dotsc, x_k-1, \dotsc, x_n).$$
The quotients $R[x_1,\dotsc, x_n]/I_k$ are ...
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M.Artin,Algebra,Chapter 12 M.10
I solve it. Seem $\mathbb C[x,y]$ as a polynomial ring over $\mathbb C[y]$.
Then, there exists $p,q,r$ belonging to $\mathbb C[y]$ such that $pf+qg=r$.
Also, there exists $p',q',r'$ belonging to $\...
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If $R$ is a Noetherian ring, then $R^n$ is Noetherian
Your observation that the ideals of $R^n$ do not split componentwise is not correct. I asked about it in this question.
Here's a way to show it using the definition of a Noetherian ring.
A ...
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Do functions over a ring have an odd and an even part?
It is not always possible to find such a decomposition.
If $f(x)=E(x)+O(x)$ where $E$ is even and $O$ is odd, then $f(x)+f(-x)=2E(x)$ is always divisible by $2$. So for instance if $R=\mathbb{Z}$ ...
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Cancellative property in integral domains
The real numbers are a field, and therefore an integral domain, and yet
$$
(0,\infty)\cdot (1,\infty)=(0,\infty)\cdot (0,\infty)=(0,\infty) \, .
$$
In fact, any integral domain $R$ which satisfies the ...
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Cancellative property in integral domains
We should definitely not expect this cancellation property of set multiplication. It's not even true in the real numbers: take for example $J=\{1\}$ and $K=\{-1,1\}$ and $I=[-7,7]$ (or any set that's ...
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If $R$ is a Noetherian ring, then $R^n$ is Noetherian
Being "Noetherian" has two meanings here. (i) $R^n$ is Noetherian as a ring; (ii) $R^n$ is Noetherian as an $R$ module. If you want to show (i), you can use that any ideal $I$ of $R^n$ is ...
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Examples of rings where every left ideal is two-sided but not every right ideal
Is there an example of a ring where every left-ideal is two-sided but not every right ideal?
This is called being a left-not-right duo ring.
Consider the twisted polynomial ring $k[x;\sigma]/(x^2)$ ...
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the center of a simple ring is either $0$ or a field
Let $a$ in $\mathbb {Z(R)}$. Then $aR$ is two sided ideal of $\mathbb {R}$. So, $aR=R$. Hence a is invertible in $\mathbb {R}$, that is $ar=1$ for some $r$ in $\mathbb {R}$. Also, $r \in \mathbb {Z(...
3
votes
Accepted
Trouble with 'elementary questions' as a beginner math student (Soft Question)
Rather than trying to give general suggestions (for this, I recommend Polya's How to Solve it), I'll go through your first example in a way that illustrates the kind of insights you (ideally) should ...
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Adjoining a function to a ring: what is this called?
Interesting question. Let's use $R \mapsto R\{f\}$ to denote this construction.
Note that there is a function $R\{f\} \to R\{f\}$ defined by $r \mapsto f(r)$. By abuse of notation, we'll also call ...
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Accepted
How to show $(k[x,y,z]/(xz,yz))_z\cong k[z]_z$?
$$\begin{align}\left(k[x,y,z]/(xz,yz)\right)_z&\cong k[x,y,z,t]/(xz,yz,1-zt)\\&\cong k[x,y,z,t]/(x,y,1-zt)\\&\cong k[z,t]/(x,y,1-zt)\\&\cong \left(k[z]\right)_z.\end{align}$$
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Small submodules of direct sum of two modules
By assumption, for some $a_i\in \mathrm{ann}(M_i)$ we have $a_1+a_2=1$, so $N=a_1N+a_2N$. Let $K_1=a_2N,\ K_2=a_1N$. It is easy to see that $K_i\subset M_i$ and $K_1\oplus K_2=N$.
Moreover, if $a_2N+L=...
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Accepted
Small submodules of direct sum of two modules
Fix $r\in \operatorname{ann}(M_1)$ and $s\in\operatorname{ann}(M_2)$ such that $r+s=1$, and let $K_1=sN$ and $K_2=rN$. Since $r+s=1$, $K_1+K_2=N$, and $K_1\subseteq M_1$ since $s$ annihilates $M_2$ ...
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Accepted
How does partial fraction expansion generalize to fractions of integers? Why is it not unique, in that case?
For relatively prime $m$ and $n$, we can split up an fraction with denominator $mn$ into fractions with denominators $m$ and $n$: $mx+ny = 1$ for some integers $m$ and $n$, so
$$
\frac{1}{mn} = \frac{...
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Fraction field of $\mathbb Z_p[[X]]$
I wanted to post this as a comment, but I don't have enough reputation. The accepted answer is wrong. It is not true that a nonzero element of $\mathbb{Z}_p[[X]]$ is of the form $X^np^m\sum_k b_kX^k$ ...
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Accepted
Quaternion algebras Brauer equivalent iff isomorphic
In fact the following is true if general:
Claim. Two central simple $k$-algebras of same degree are Brauer equivalent if and only if they are isomorphic.
Note that Brauer equivalence for two arbitrary ...
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