# Tag Info

## New answers tagged abstract-algebra

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### Problem with understanding why $\mathbb{C}[x, y] / (x^2 + y^2 - 1)$ is UFD

There is an isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ given by $T^{\pm1}=X\pm iY$. With this identification in mind, we see $X=(T+T^{-1})/2=\frac{1}{2}T^{-1}(T+i)(T-i)$ is not ...
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### In any ring $R$:product of non-unit and any element is non unit.

Consider the abelian group $A=\Bbb Z^{\omega}$ of sequences of integers and let $R=\operatorname{End}(A)$ be its ring of endomorphisms. For $f\in A$, define $D(f)$ per $$D(f)(n)=f(n+1)$$ and $I(f)$ ...
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### exact sequence of differential

$N$ is not just any $B$-module, it’s a $B$-module where the $B$-action factorizes through $C$ – which means exactly that $I$ annihilates $N$.
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Here is an alternative answer to this old question. The key point is to show the following: Given a subgroup $H$ of a group $G$, there is a group $K$ and homomorphisms $a, b:G\to K$ such that $a(g)=b(... • 1,186 1 vote Accepted ### Prove$ \bar{u}\in \operatorname{Hom}_{A} (\operatorname{Hom}(M,N), \operatorname{Hom}(M',N))$Suppose$u : M' \to M$is an$A$-module homomorphism. Then$\overline{u} : \text{Hom}(M, N) \to \text{Hom}(M', N)$is defined by $$\overline{u}(f) = f \circ u.$$ First, observe that$f \circ u$is an$...
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here are the two missing elements: $$(A \cup B) \cap (A^c \cup B^c), (A \cap B) \cup (A^c \cap B^c)$$ These are complementary each other. The first one is the symmetric defference of $A,B$, while the ...
For instance, the lexicographically ordered group $(\mathbb{Z}^2,+,<)$ is the underlying ordered group of no ordered ring. edit: in fact, what I show below is rather that there is no structure of ...