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My idea is that using Classification of prime ideals of $\mathbb{Z}[X]$

I considered $(x,p)$ where $p>6$ is prime in $\mathbb Z$

or type of $(x^n)$ or $(p)$, but above is not contained in the given ideal and these last two are not prime or again not contained in $(6,x)$

Maybe some hint, answer would be much appreciated.

Micheal Brain Hurts
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    Hint: if $p(x)$ is a polynomial whose constant coefficient is divisible by $6$, then $p(x)\in (x,6)$. If $p(x)$ is irreducible, then $(p(x))$ is prime. – Arturo Magidin Dec 20 '22 at 20:35

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Find an irreducible polynomial $f(x)$ of the form $6a(x) + xb(x)$. Then $(f(x))$ is a prime ideal contained in $(6,x)$. There are many such irreducibles of each degree.

J. W. Tanner
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KCd
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