115 reputation
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age 20
visits member for 2 years
seen Nov 16 '13 at 18:22

Jul
2
awarded  Curious
Dec
26
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 $x$?
Dec
25
comment (Ir)reducible polynomial over $\mathbb{Z}$
ok, sorry anyone
Dec
25
revised (Ir)reducible polynomial over $\mathbb{Z}$
deleted 3 characters in body
Dec
25
asked (Ir)reducible polynomial over $\mathbb{Z}$
Dec
25
comment Perfect Square in an UFD
Can you give me a counterexample
Dec
25
accepted Perfect Square in an UFD
Dec
25
asked Perfect Square in an UFD
Dec
16
comment About Classification of Finitely Generated Abelian Groups
Yes, I did. Now can anyone help me?
Dec
16
awarded  Scholar
Dec
16
awarded  Supporter
Dec
16
accepted Prove that exists a unique subgroup $H$ of $G$ has order of $n$.
Dec
16
accepted $G = \mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ contains a element of order $m$ iff $m\mid n_1$.
Dec
16
accepted $(\mathbb{Q},+)$ has no maximal subgroup
Dec
16
asked About Classification of Finitely Generated Abelian Groups
Dec
5
asked $G = \mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ contains a element of order $m$ iff $m\mid n_1$.
Nov
22
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.
Nov
21
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.
Nov
20
asked $F[x]/\langle f(x) \rangle$ has $q^n$ elements
Nov
15
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.