# Firmino

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bio website location age 20 member for 2 years, 2 months seen Nov 16 '13 at 18:22 profile views 163

# 64 Actions

 Jul2 awarded Curious Dec26 comment (Ir)reducible polynomial over $\mathbb{Z}$ I can't apply Eisenstein's criterion because coefficient of $x^3$ is $1$. So my strategies is shift $x$. But power of $x$ is very big. How should I shift $x$? Dec25 comment (Ir)reducible polynomial over $\mathbb{Z}$ ok, sorry anyone Dec25 revised (Ir)reducible polynomial over $\mathbb{Z}$ deleted 3 characters in body Dec25 asked (Ir)reducible polynomial over $\mathbb{Z}$ Dec25 comment Perfect Square in an UFD Can you give me a counterexample Dec25 accepted Perfect Square in an UFD Dec25 asked Perfect Square in an UFD Dec16 comment About Classification of Finitely Generated Abelian Groups Yes, I did. Now can anyone help me? Dec16 awarded Scholar Dec16 awarded Supporter Dec16 accepted Prove that exists a unique subgroup $H$ of $G$ has order of $n$. Dec16 accepted $G = \mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ contains a element of order $m$ iff $m\mid n_1$. Dec16 accepted $(\mathbb{Q},+)$ has no maximal subgroup Dec16 asked About Classification of Finitely Generated Abelian Groups Dec5 asked $G = \mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ contains a element of order $m$ iff $m\mid n_1$. Nov22 comment $(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$ It is amazing. Thank you very much. Nov21 comment $(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$ I think isomorphism: $$\varphi : f(x)+\langle x^2 \rangle \mapsto f(x+1)+\langle (x+1)^2 \rangle$$. This is 1-1 mapping and homomorphism. Nov20 asked $F[x]/\langle f(x) \rangle$ has $q^n$ elements Nov15 comment $(\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -2\rangle \ncong (\mathbb{Z}/2\mathbb{Z})[x] /\langle x^2 -3\rangle$ I think everyone know definition $\mathbb{Z}_2$, so I don't need confirm what its mean.