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I'm just getting started learning ring theory and am currently learning about ideals. By book (Dummit & Foote) says the following:

For example, in the ring $R = \mathbb{Z}[x]$ the elements $2$ and $x$ generate a maximal, non-principal ideal

First off, for clarification, the notation $\mathbb{Z}[x]$ just means "the set of all polynomials with integer coefficients," right (i.e. the set of all

$$ a_0 + a_1x + a_2x^2 + \cdots $$

such that each $a_i \in \mathbb{Z}$)? Second, how is the ideal generated by $2$ and $x$ maximal? If you let $I$ be the ideal generated by $2$ and $x$ and let $J$ be the ideal generated by $1$ and $x$, isn't it true that $I \subset J$ and so $I$ is not maximal?

Any clarification would be greatly appreciated!

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    $\begingroup$ $J=R$ is the whole ring, so there is no problem. $I$ is a maximal ideal. $\endgroup$ – Dietrich Burde Dec 13 '14 at 21:15
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You are correct about the definition of $\mathbb{Z}[x]$.

Maximal in this context means "maximal among proper ideals". The ideal $J$ that you suggested is in fact all of $\mathbb{Z}[x]$.

To see that $I$ is maximal among proper ideals, suppose that another ideal $K$ contains $I$. Since $I$ is the set of polynomials with even constant term, either $K=I$ or $K$ contains a polynomial with odd constant term, $a_nx^n+\dots+a_1x+2k+1$, in which case, we may subtract off the polynomial $a_nx^n+\dots+a_1x+2k\in I\subset K$ to see that $1\in K$ and hence $K=\mathbb{Z}[x]$.

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  • $\begingroup$ More to the point: since $1$ is invertible in $\Bbb{Z}[x]$, the ideal generated by $1$ is all of $\Bbb{Z}[x]$. $\endgroup$ – Cameron Williams Dec 13 '14 at 21:16
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An ideal $I$ in $R$ is maximal if the quotient $R/I$ is a field. In our case, the quotient is the field $\mathbb{F}_2$, hence $I$ is maximal. For a proof see also here. The ideal $I$ is not a principal ideal by degree considerations.

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