I am currently learning abstract algebra, and I am going over prime elements of a ring. I am a little confused as to the connection between the definitions of prime numbers and prime elements. A prime number is an integer $p$ greater than $1$ such that $a\mid p \Rightarrow (a=1 \vee a=p)$ for an integer $a$ greater than $0$.
A prime element is an element $p$ of a commutative ring with identity such that for $x,y\in R$, $p\mid xy \Rightarrow (p\mid x \vee p\mid y)$.
Now, I know that a prime element is more general than a prime number. I just want to prove that if $p$ is a prime number in the integers, then it satisfies the prime element property. Can anyone provide a proof? I'm completely stuck.
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Consider the unique factorization of $x$ and $y$ into prime factors. – Stefan Mesken Feb 29 '16 at 00:38
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If a prime $p$ divides a product $ab$ in the integers, what do we know? Then $p$ must be in the prime factorization of either $a$ or $b$, since we can't decompose $p$ into a product of 2 nonunit integers. A prime element in a ring has the same idea behind it, and is inspired by the integer ring $\mathbb{Z}$. – smingerson Feb 29 '16 at 00:40
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If $p$ does not divide $x$, then $p$ and $x$ are relatively prime. It follows (Bezout) that there exist integers $s$ and $t$ such that $sx+pt=1$. Multiply by $y$. We get $sxy+pty=y$. Note that $p$ divides both terms on the left-hand side, so $p$ divides $y$. The problem has been solved repeatedly on MSE. – André Nicolas Feb 29 '16 at 00:43