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reviewed No Action Needed Proof of 2^n deck of card, it will be reverse order performing n perfect in-shuffle.
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comment Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$.
In an integral domain $R$, $r,s\in R$ are associates iff $r=su$ for some unit $u\in R$. If $r$ divides $z$, $rs=z$ for some $s$. By my definition, if $z$ is irreducible, either $r$ or $s$ is a unit. So $r|z$ implies $r$ is a unit or $r$ is an associate of $z$.
Mar
14
comment Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$.
For a ring $R$, $r\in R$ is a unit if $\exists s\in R$ such that $rs=sr=1$. An element in a ring is irreducible if it is not the product of two nonunits.
Mar
14
answered Prove that if $N(z)$ is irreducible in $\Bbb{Z}$ then $z$ is irreducible in $\Bbb{Z}[\sqrt{\alpha}]$.
Mar
13
reviewed No Action Needed Question about noise term in SDEs