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Artinian ring with zero finitistic dimension

In the page p178 of the paper PREDICTING SYZYGIES OVER MONOMIAL RELATIONS ALGEBRAS says the following. When $\Lambda$ is a left artinian ring, the following are equivalent: $\operatorname{fin\,dim} \...
Liang Chen's user avatar
0 votes

What is the norm of an ideal $(2,1+\sqrt{-5})$ in $\mathbb Z[\sqrt{-5}]$?

You can use Hermite normal form, like Bill Dubuque suggests. However, you have two alternative ways to proceed. (1) You can show that $(2)=I\bar{I}$, then use multiplicativity of the norm and the fact ...
GreginGre's user avatar
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7 votes
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If $R/\langle r\rangle\cong k[x_1,\dots, x_n]$, does this implies that $R\cong k[x_1,\dots, x_n,x_{n+1}]$?

Consider the localization $k[x]_{(x)}$ this has only one nonzero prime ideal so it's definitely not isomorphic to $k[x]$ which has infinitely many prime ideals. Note that we have $k[x]_{(x)}/(x) \cong ...
Lukas Heger's user avatar
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0 votes

Is every vector space an injective module?

For any vector space $M$, $$\mathrm{Ext}^1_F(M,Q)=0$$ since $M$ is free, hence projective. Thus, $Q$ is injective.
Juan L.'s user avatar
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4 votes
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Prime radical of a ring

It shouldn't read "Suppose $I$ is a prime ideal, then find their intersection." Rather it should read "Find all prime ideals of $R$, then find their intersection." So, you can see ...
rschwieb's user avatar
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1 vote

Generators and presentations of a division ring

When defining a group presentation, you work in the category of groups, so you want to force your construction to have inverses. When defining an algebra presentation, you work in the category of ...
Captain Lama's user avatar
3 votes

For which rings $P, Q, P\times Q$ product is a field? - Abstract Algebra problem from my college exam.

In short, you are right to look at elements of the form $(p,0)$, but you seem to be jumping to conclusions. Instead, start from the definitions and deduce something about $P\times Q$. Suppose $(p,0) ...
Brian Moehring's user avatar
0 votes

Showing that z is a root of $X^n -1$ iff z is an element of the set of generators of dth roots of unity for some $d\mid n.$

You're off a little, but playing with some of the tools. Suppose $z^n=1$ and $z$ is not a primitive $n$th root of unity. Then $z=e^{2\pi ik/n},$ where $(k,n)\gt1.$ But then $z$ is a primitive $n/(n,k)...
soft tostada burrito's user avatar
2 votes

What ring homomorphisms do induce well-defined mapping between principal ideals?

You don't specify if your rings are commutative or have unities, so let me do this as generally as possible for as long as possible... Let $R$ and $S$ be ring, and let $\phi\colon R\to S$ be a ring ...
Arturo Magidin's user avatar
1 vote
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Kernel of a derivation over an arbitrary ring

If it's "linear with respect to addition", it's at least an abelian group homomorphism, hence $D(0)=0$ and in fact $\ker D$ is an additive subgroup. And if $D(x)=D(y)=0$, surely $D(xy) = xD(...
Just a user's user avatar
3 votes

Proof that ring of rational upper triangular matrices is a right Goldie ring

Since it is a finite dimensional $\mathbb Q$ algebra, it is also right and left Artinian, and right and left Noetherian. The fact that it's Noetherian makes both of the Goldie conditions hold ...
rschwieb's user avatar
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1 vote

If p is prime then J = $\left \langle p, a + \sqrt{D} \right \rangle_{\mathbb{Z}}$ is ideal prime.

Your description of $J$ agrees with the more concise description $J = {\mathbf Z}p + \mathbf Z(a+\sqrt{D})$: it's the $\mathbf Z$-linear combinations of $p$ and $a+\sqrt{D}$. You should first prove ...
KCd's user avatar
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0 votes

Radical of an ideal equals the intersection of all prime ideals containing it.

Let $x\in\sqrt{I}$. Then $x^n\in I$. Pick an arbitrary $\mathfrak{p}$ containing $I$. Then $x^n\in I\subset\mathfrak{p}\implies x\in\mathfrak{p}$. For the other inclusion, suppose for the sake of ...
Harry A's user avatar
0 votes

Describe all ring homomorphisms from $\mathbb Z\times\mathbb Z$ to $\mathbb Z\times\mathbb Z$

Any ring homomorphism needs to be a group homomorphism on the additive group. By a theorem of Dedekind, every subgroup of a free abelian group of rank $n$ is free abelian, of rank less than or equal ...
soft tostada burrito's user avatar
0 votes

Describe all ring homomorphisms from $\mathbb Z\times\mathbb Z$ to $\mathbb Z\times\mathbb Z$

By CRT, $$\mathbb Z[x]/\langle x(x-1)\rangle \simeq \mathbb Z[x]/\langle x \rangle \times \mathbb Z[x]/\langle x-1\rangle \simeq \mathbb Z\times\mathbb Z$$ Therefore by the universal property of the ...
Just a user's user avatar
0 votes

Describe all ring homomorphisms from $\mathbb Z\times\mathbb Z$ to $\mathbb Z\times\mathbb Z$

I will follow Jordan's idea and give a detailed answer. He has shown that $f(1,0)$ has four possibilities:$(1,0),(0,1),(1,1),(0,0)$. Similarly, assume $f(0,1)=(m,n)$, $f((0,1)(0,1))=f(0,1)f(0,1)=(m^2,...
Morten's user avatar
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0 votes
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Ring homomorphism may not preserve $1$.

I believe a good "reason" can come from a good example. The main idea though (I think) is that just because the image of the identity needs to act like the identity (due to homomorphism ...
Kyle's user avatar
  • 26
0 votes

Demonstrate that a set of 2x2 matrices form a finite commutative ring of unity

This seemed strange enough to me that I pulled up the book. The quoted text is exactly correct: Compare a purported ring structure on this set of four matrices to the ring $\mathbb{Z}/4\mathbb{Z}$. I ...
bob's user avatar
  • 11
3 votes
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If $R$ has only a finite number of simple modules up to isomorphism, does it necessarily follow that $R$ is a semi-local ring?

There might be much simpler examples, but I found the following example on the DaRT Database of Ring Theory. It's a bit complicated to define from scratch (it depends on Kolchin's definition of a ...
Jeremy Rickard's user avatar
1 vote
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Euclidean algorithm in commutative rings with unity

Your intuition seems right, but it may be easier to use what we have in $\mathbb{Z}$ rather than try to copy-paste it onto a new ring. This will avoid concerns about new factors (e.g. presence of ...
While I Am's user avatar
  • 2,444
2 votes

Proving a module is injective (or not) and failed attempt using Baer's Criterion.

Again I prefer to use $F_2$ for clarity and ease of typing. One thing that simplifies the situation here is how many summands of $R$ there are. Each one that is a direct summand of $_RR$ easily ...
rschwieb's user avatar
  • 154k
1 vote

Proving a module is injective (or not) and failed attempt using Baer's Criterion.

$\require{AMScd}$ This problem can be easily solved by using the language of quiver representations for finite-dimensional algebras over a field. If you are not familiar with this theory, it is a very ...
Javier Herrero's user avatar
1 vote

Is every sufficiently large gaussian integer the sum of 3 cubes ? $a + b i = (c + di)^3 + (e + fi)^3 + (g + hi)^3$?

I can answer the second question . . . Let $\omega$ be a primitive cube root of unity. Claim:$\;$For integers $a,b$ with $a$ odd and $b\equiv 2\;(\text{mod}\;4)$, there are no integers $c,d,e,f$ for ...
quasi's user avatar
  • 59k
2 votes

$H \leqslant (\Bbb{Z}/n)^*$ the square roots of $1$, with $n$ squarefree; what's size of pseudocoset $kH\subset \Bbb{Z}/n$ where $k$ is any integer

Let $\gcd(k,n) = d$. Then by the Orbit-Stabilizer theorem we have that $|kH| = |\text{Orb}(k)| =|H|/\text{Stab}(k)$ where $\text{Stab}(k) = \{ h \in H : hk = k \pmod n\}$. One such example is $11\...
Daniel Donnelly's user avatar
1 vote
Accepted

$H \leqslant (\Bbb{Z}/n)^*$ the square roots of $1$, with $n$ squarefree; what's size of pseudocoset $kH\subset \Bbb{Z}/n$ where $k$ is any integer

The size of $H$ depends on the number of prime factors of $n$. If $n=p_1\cdots p_r$, with $p_1\lt p_2\lt\cdots\lt p_r$ distinct primes, then $|H|=2^r$ if $p_1\neq 2$, and $|H|=2^{r-1}$ if $p_1=2$. In ...
Arturo Magidin's user avatar
0 votes

Primes of the form $x^2 + y^2$ can only be written as $a^2 + b^2$ in one way?

Look at https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares Euler gave an elementary proof.
mick's user avatar
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1 vote

Matrices over an integral domain

Most facts that are true for matrices over a characteristic-zero field like $\mathbb{R}$ or $\mathbb{Q}$ are going to also be true about matrices over any field $F$ of characteristic $p > 0$. For ...
Joppy's user avatar
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1 vote
Accepted

Finding the projective cover of a quotient of ring matrices.

It turns out that $I$ is a projective left $R$-module, but $R/I$ isn't. Denote ($*$ stands for an arbitrary element of $\Bbb{Z}_2$). $$ J_1=\{\left(\begin{array}{cc}*&0\\ 0&0\end{array}\right)...
Jyrki Lahtonen's user avatar
1 vote

Show that $\text{Frac}(\mathbb{Z}(\sqrt{2}))$ is isomorphic to $\mathbb{Q}(\sqrt{2})$

Take $\frac{p}{q} + \frac{r}{s}\sqrt2 \in \mathbb{Q}[\sqrt2]$. Then we have $\frac{p}{q} + \frac{r}{s}\sqrt2 = \frac{ps+qr\sqrt2}{qs+0\sqrt2} = \alpha([ps+qr\sqrt2,qs+0\sqrt2])$. Since $q,s \not= 0$ ...
Jeremy Davie's user avatar
0 votes

Reducible and indecomposable modules/representations

A module $M$ is reducible if there is a nontrivial submodule $N$, i.e., $0 \subset N \subset M$, where the inclusions are proper. But there are lots of examples of such submodules that do not have a ...
Sammy Black's user avatar
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0 votes

Reducible and indecomposable modules/representations

None of this is special to modules over a group ring, so let's just assume we have some ring $A$ and a left module $M$. $M$ is called simple or irreducible if there are exactly two submodules (which ...
Lukas Heger's user avatar
  • 21.6k
1 vote
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A Question Concerning the Properties of a Projective Cover

Let $R$ be the triangular matrix ring $\begin{pmatrix}\mathbb{Z}/2\mathbb{Z}&\mathbb{Z}/2\mathbb{Z}\\0&\mathbb{Z}\end{pmatrix}$, and let $P$ be the projective left module $\begin{pmatrix}\...
Jeremy Rickard's user avatar
1 vote
Accepted

Koszul relations of a regular sequence $x_ix_j-x_jx_i=0$

In general a Kosuzul complex of the pair $(\mathbf{x},M)$, where $\textbf{x} = \{x_1,...,x_n\}$ is an $R$-sequence (not necessarily $M$-regular) and $M$ is an $R$-module, is a chain complex $K_{\...
RHspqr's user avatar
  • 362
1 vote

Associated graded ring of a quotient

$\newcommand{\fraka}{\mathfrak{a}} \newcommand{\frakb}{\mathfrak{b}} \newcommand{\frakm}{\mathfrak{m}} \DeclareMathOperator{\Gr}{Gr}$Let $\fraka^*$ denote the ideal of $\Gr_I(A)$ generated by the ...
mbert's user avatar
  • 259
0 votes

There is, up to isomorphism, a unique smallest field $\mathbb{Q}$, which contains $\mathbb{Z}$ as a subring

This looks correct to me, except for the definition of $h$ in Step 2. $\Bbb Q$ was defined as the quotient set $\Bbb Z \times \Bbb Z^\times / \sim$ with $\sim$ the equivalence relation given in the ...
EE18's user avatar
  • 1,143
1 vote
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How many zero divisors in $\Bbb Z_5[x]/\langle x^3+2x+2 \rangle$

The variant/application of the first isomorphism theorem you want is actually the Chinese remainder theorem. Let $R=\mathbb Z_5[x]$, $I=((x-1)^2)$, and $J=(x-3)$ Consider the map $R\to R/I \times R/...
Aaron's user avatar
  • 24.3k
2 votes

How many zero divisors in $\Bbb Z_5[x]/\langle x^3+2x+2 \rangle$

I suppose by $\mathbb{Z}_5$ you mean $\mathbb{Z}/5\mathbb{Z}$. As $x^3+2x+2=(x-1)^2(x-3)$ and the (principal) ideals $(x-1)^2$ and $(x-3)$ are coprime, Chinese Remainder Theorem implies \begin{align*}\...
Schlitzer's user avatar
  • 114
0 votes

How is every cyclic submodule of $\mathbb{Q_{Z}}$ is a small submodule?

Here is a more general conclusion. Because $ _{\mathbb{Z}}\mathbb{Q} $ does not contain any maximal submodules, the following conclusion can be applied to your problem: Consider a ring $ R $. If a ...
Liang Chen's user avatar
1 vote

Trying to find a subring of $M_4(\mathbb{R})$ isomorphic to $\mathbb{H}$, the quaternions

Define the function $\Phi : \mathbb{H} \to M_4(\mathbb{R})$ by mapping a quaternion $ a + bi + cj + dk $ to a $4 \times 4$ matrix: $$\Phi(a + bi + cj + dk) = \begin{pmatrix} a & b & c & d \...
Liang Chen's user avatar
1 vote

counterexample that $M$ is not finitely generated $R$-module and $M$ has no maximal submodule

$ \mathbb{Q} $ as a $ \mathbb{Z} $-module is not finitely generated and $_{\mathbb{Z}}\mathbb{Q}$ has no maximal submodule. To show this, assume for contradiction that $ \mathbb{Q} $ is finitely ...
Liang Chen's user avatar
1 vote

To show an isomorphism is valid, is it sufficient to just show that the order you apply it is irrelevant?

Showing $\phi$ respects addition and multiplication only shows that it is a homomorphism. To show it is an isomorphism, you need to show that it is a bijection. One way to do that is to define a $\psi$...
ultralegend5385's user avatar
0 votes

Ring theory exercises at the graduate level

One that sounds interesting is Rings and Their Modules, by Paul E. Bland (University of Kentucky). The MAA review starts it's "a welcome addition to the theory of rings". Supposedly it has ...
soft tostada burrito's user avatar
1 vote

Technical question about computing the left ideals of the ring $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$

It seems to me you are not doing step 3 clearly. We are agreed on step 1. I'm going to use $F_2$ since it is easier to type and less ambiguous than $\mathbb Z_2$. Then in step 3, there are two cases: ...
rschwieb's user avatar
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2 votes
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Is there a non-commutative ring with unity $R$ such that its Jacobson radical is not a two-sided maximal ideal?

The Jacobson radical is not often maximal, because $R/J(R)$ does not have to be a simple ring. It would seem you have answered your own question already, because your example is fine. Given any two ...
rschwieb's user avatar
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1 vote
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Computing the projective cover of a module over a ring of matrices.

We have $R \cong S \oplus S'$ as you described in the comments. This implies $R/S \cong S'$ is projective, as it is a direct summand of a free module.
Daniel Arreola's user avatar
0 votes

Questions about correct notations used in showing an ideal is a maximal ideal.

Lots of problems here. The set $\{a+b\sqrt[n]{D}\mid a,b\in\mathbb{Z}\}$ is not a ring unless $(\sqrt[n]{D})^2$ can also be expressed this way (otherwise it is not closed under multiplication), and ...
Arturo Magidin's user avatar
1 vote
Accepted

Annihilator basis after extension of scalars

Note that $S_a$ is the kernel of the map $a : A \to A \oplus A$ given $$ x \mapsto (ax, xa). $$ Thus, we have an exact sequence $$ 0 \to S_a \to A \xrightarrow{a} A \oplus A. $$ By exactness of the ...
Frank's user avatar
  • 2,603
1 vote
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Proof or counterexample of the inclusion $\mathrm{Ass}(I)\subset\mathrm{Ass}(I^N)$ for monomial ideals

Here is a counterexample taken from [1, Example 4.18]: Take $R=k[a,b,c,d,e]$ and $I=(ab^2c,bc^2d,cd^2e,de^2a,ea^2b)$. One can compute (using Macaulay2 for instance) that $(a,b,c,d,e)$ is an associated ...
cqfd's user avatar
  • 12.3k
3 votes

Technical question about computing the left ideals of the ring $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$

It is not true that the ring $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$ is "basically the matrix $\left(\begin{array}{cc} \mathbb{Z}_2&\mathbb{Z}_2\\ 0&\mathbb{Z}_2\end{array}\...
Arturo Magidin's user avatar
1 vote
Accepted

If $f: A \to B$ is a homomorphism, then a prime $\mathfrak{p} \subseteq A$ is the contraction of a prime in $B$ iff $\mathfrak{p}^{ec} = \mathfrak{p}$

Assume that $\mathfrak{p}^{ec} = \mathfrak{p}$. We need to show that there is a prime that contracts to $\mathfrak{p}$. As the OP pointed out, $\mathfrak{p}^e$ may not be prime in $B$. We will find a ...
Agilan's user avatar
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