# Tagged Questions

This tag is for questions about rings, which are a type of algebraic structure studied in abstract algebra and algebraic number theory.

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### How can I prove that something is an irreducible element?

I want to show that $4 + i$ is an irreducible element of $\mathbb{Z}[i]$. My current approach is to let $4+i$ = $AB$, where $A$ and $B$ are elements of $\mathbb{Z}[i]$, where $A=a+ia'$ and ...
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### Does $|x^*|=|x|$ in a star ring with an absolute value?

Let $R$ be a star ring with an absolute value. Is it true that $|x^*|=|x|$ for all $x\in R$? Here a star ring is a ring with a function $*:R\to R$ called conjugation such that $(x+y)^*=x^*+y^*$ ...
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### The units of $(\mathbb{Z}/4\mathbb{Z})[x]$

I want to find the units of $(\mathbb{Z}/4\mathbb{Z})[x]$, that is, the units of the set of polynomials of x modulo 4. Where can I start with this?
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### proof that $\mathbb Z_p/p^n\mathbb Z_p\cong \mathbb Z/p^n\mathbb Z$

I'm trying to prove the isomorphism showed in the title. By the theory of valuations I have a succession of ideals: $$\mathbb Z_p\supset p\mathbb Z_p\supset\ldots\supset p^n\mathbb Z_p\supset\ldots$$ ...
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### $|x^{-1}-y^{-1}|=|x-y|/|x||y|$ in a normed ring

I hit a slight snag when trying to prove that the inverse function $x\mapsto x^{-1}$ on the unit group is continuous in a ring with an absolute value, so I'd like some confirmation that the theorem is ...
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### Is “polynomials in $x$” a monad?

The construction of polynomials $R \mapsto R[x]$ gives a functor $P: \mathbf{Ring} \to \mathbf{Ring}$ on the category of possibly noncommutative rings. Choosing a ring $R$ for the moment, there is a ...
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### Show that $\operatorname{rad}(I)=\bigcap P$ for all $P$ prime containing $I$

Let $R$ be a commutative ring with identity and let $I$ be an ideal of $R$. Define $\operatorname{Rad}(I)=\{a\in R:\exists n\in\mathbb N, a^n\in I\}$. Show that $Rad(I)$ is the intersection of all ...
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### Permutation groups acting on polynomial rings and base change

Let $G\subset S_n$ be a permutation group, and let it act on $R = \mathbb{Z}[x_1,\dots,x_n]$ by permuting the variables, as usual. $G$ acts on $R\otimes_\mathbb{Z} S$ for any unital ring $S$ via its ...
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### The homomorphism $h:(J:I)\rightarrow \text{Hom}_R(R/I,R/J)$ is surjective.

We've got this homomorphism $$h:(J:I)\rightarrow \text{Hom}_R(R/I, R/J)$$ $$x\mapsto h_x$$ where $$h_x:R/I\rightarrow R/J$$ $$r+I\mapsto xr+J.$$ We found a mistake in the proof of the ...
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### Prove that a mapping from C to M2(R) is injective and a homomorphismm

$\mathcal{M}_2(\Bbb R)$ is the ring of 2x2 matrices with real entries. Define a map $\phi:\Bbb C \to \mathcal{M}_2(\Bbb R)$ by $$\phi(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$ ...
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### Factorizing $(11 - \sqrt{-14})$ in $\mathbb{Z}[\sqrt{-14}]$ as product of maximal ideals

For an exercise I would like to: Factorize $(11 - \sqrt{-14})$ in $\mathbb{Z}[\sqrt{-14}]$ as a product of maximal ideals which is possible as it is a Dedekind domain (it is the ring of integers ...
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### Rank of a matrix over a principal ideal domain

I apologize if my question is stupid but I'm not very familiar with matrices over a principal ideal domain $R$ (For example, $R=\mathbb{Z}$ or $R=\mathbb{R}[X]$). I was wondering how to define the ...
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### About some isomorphism of right R-modules as abelian groups

Let $M$ and $V$ be right $R$-modules. Let $D=End_{R}(M)=Hom_{R}(M,M)$. Suppose that $Hom_{R}(V,Hom_{R}(M,V)\otimes_{D}M)) \cong Hom_{R}(V,V)$ as abelian groups, where this isomorphism is given by ...
### Counting monic polynomials over $\mathbb{Z}_{p^k}$ with at least $2$ distinct roots
At one point in the course of working on something I was obliged to count the number of monic polynomials of degree $n$ over $\mathbb{Z}_p$ having at least two distinct roots (as usual, $p$ is a ...
### $c$ is irreducible implies that $\langle c\rangle$ is a maximal ideal, proof verification
This is my own personal proof, it seems right, but I want to make sure I don't carry incorrect logic into my future work. Let $R$ be a PID and $c\in R$ If $c$ is irreducible, then \$\langle ...