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Let $S$ denote the set of polynomials in $x$ with integer coefficients that have no linear term. In other words, $S = \{ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_0 \,|\, n \in \mathbb{Z}^{\geqslant 0}, a_i \in \mathbb{Z} \}.$

How is $x^2$ prime in this ring? It seems like $x^2 \mid x^6$ by $x^2 \cdot x^4 = x^6$. But $x^6 = x^3 \cdot x^3$ and $x^2$ does not divide $x^3$ in this ring. Help?

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    Who said it was prime? – 2017-02-09
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    The professor said it was prime – 2017-02-09
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    They surely wanted to say *irreducible*. – 2017-02-09
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    The professor mentioned that he was wrong. Sorry for the inaccurate question. – 2017-02-12

2 Answers 2

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You're correct: $x^2$ is not a prime element of the ring $S$. Indeed, this is a standard example of an element of a domain that is irreducible but not prime.

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    Can you provide two polynomials $\notin (x^2)$ such that their product $\in (x^2)$ ? – 2017-02-12
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    The question gave such an example: $x^3$ and $x^3$. – 2017-02-12
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    Indeed $x^3 \notin (x^2)$ but its square does. – 2017-02-12
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A frequent definition of primality in rings is the following: a principal ideal $(w)$ is prime $\iff$ $\forall u,v$ with $u.v \in (w) \implies u \in (w)$ or $v \in (w)$. Note that the the polynomials of the ideal $(x^2)$ have a zero constant term .Can you write two polynomials that $\notin (x^2)$ such that their product $\in (x^2)$?

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    Have you even read the question? The OP has given such an example showing that he knows very well what's a prime element. (-1) – 2017-02-12