By definition, we have that an ideal $I$ is a prime ideal if, for all $ab \in I$, either $a \in I$ or $b \in I$. However, I don't know how to use this definition to show that $(2, X)$ is a prime ideal.
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5You may want to prove that $\mathbb{Z}[X]/I$ is a field. Hint: consider the map $\mathbb{Z}[X] \to \mathbb{Z}/2 \mathbb{Z}$ which sends $a_{0} + a_{1} X + \dots$ to the class of $a_{0}$ modulo $2$. This is a ring homomorphism. What's the kernel? – Andreas Caranti Dec 04 '16 at 19:36
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@Dieritch Burde Should I delete this then? – User203940 Dec 04 '16 at 19:45
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@Jmars Deletion is up to you. If you feel that the question and the answers were more helpful or add something new (that wasn't in the "possible duplicate" link), it is not a problem to keep this question. – Michael Burr Dec 04 '16 at 19:49
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I guess I'll leave this up since many of the hints and solutions given are easy to understand. – User203940 Dec 04 '16 at 19:52
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${\bf{Z}}[X]/\left<2,X\right>\cong{{\bf{Z}}_{2}}[X]/\left<X\right>\cong{{\bf{Z}}_{2}}$, so what can you conclude about?
user284331
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It's a field and so therefore the kernel $<2, X>$ must be a prime ideal, correct? – User203940 Dec 04 '16 at 19:42
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1@Jmars actually it more precisely means $(2,x)$ is a maximal ideal, which is in particular prime. – Alex Mathers Dec 04 '16 at 19:43
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1Right, because if it's a field it's also by extension an integral domain. – User203940 Dec 04 '16 at 19:46
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$\langle 2,X\rangle$ consists of all polynomials in $\mathbb{Z}[X]$ of the form $2f(X)+Xg(X)$, in other words, this ideal consists of all polynomials of the form $$ a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0 $$ where $a_0$ is even.
If $fg\in\langle 2,X\rangle$, then the constant term of $fg$ is even. Since the constant term of $fg$ is the product of the constant terms of $f$ and $g$, one of $f$ or $g$ is in the ideal. Hence, the ideal is prime.
Michael Burr
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