Question about an example on ring theory from Dummit and Foote

Background

Example: If $$p$$ is a prime, the ring $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$ obtained by reducing $$\Bbb{Z}[x]$$ modulo the prime ideal $$(p)$$ is a Principal Ideal Domain, since the coeffiencets lie in the field $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$. This example shows that the quotient of a ring which is not a Principal Ideal Domain may be a Principal Ideal Domain. To follow the ideal $$(2,x)$$ above in this example, note that if $$p=2$$, then the ideal $$(2,x)$$ reduces to the ideal $$(x)$$ in the quotient $$\Bbb{Z}[x]/2\Bbb{Z}[x]$$, which is a proper (maximal) ideal. If $$p\neq 2$$, then $$2$$ is a unit in the quotient, so the ideal $$(2,x)$$ reduces to the entire ring $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$.

Questions

For above example, I am having a lot of trouble parsing the sentences. Before anything, I want to say that I understand why the ideal $$I$$ consisting of polynomials with even constant terms is not a principal ideal in $$\Bbb{Z}[x]$$, and that the the meaning of the following notaations: $$\Bbb{Z}[x]/2\Bbb{Z}[x]:=\Bbb{Z}[x]/(2\Bbb{Z}[x]), \Bbb{Z}[x]/p\Bbb{Z}[x]:=\Bbb{Z}[x]/(p\Bbb{Z}[x])$$

The elements of $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$ are of the form $$f(x)+ \langle p\Bbb{Z}[x] \rangle$$. I am not sure why the coeeficients lie in the field $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$ would make a difference. If $$p$$ is prime, what do the phrase: "....reducing $$\Bbb{Z}[x]$$ modulo the prime ideal $$(p)$$ is a Principal Ideal Domain" mean in math notation.

Next the sentence: "if $$p=2$$, then the ideal $$(2,x)$$ reduces to the ideal $$(x)$$ in the quotient $$\Bbb{Z}[x]/2\Bbb{Z}[x]$$, which is a proper (maximal) ideal", does it mean since $$(2,x)=(2)+(x)$$, then $$(2)+(x)\equiv (x) \pmod {(2x)}$$? Also, what is the differnece between a maximal ideal versus a proper maximal ideal?

Lastly, I am having trouble understanding the last phrase: "If $$p\neq 2$$, then $$2$$ is a unit in the quotient, so the ideal $$(2,x)$$ reduces to the entire ring $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$." why is 2 a unit in the quotient and also why then would the ideal $$(2,x)$$ reduces to the entire ring $$\Bbb{Z}[x]/p\Bbb{Z}[x]$$?

• First paragraph of questions, what do you mean by "the ideal $I$ consisting of polynomials with constant terms" ? Commented Jul 15 at 10:21
• @FançoisGatine i edited my post, I meant even constant terms. Thank you for noticing.
– Seth
Commented Jul 15 at 10:25

For your first question, I believe the authors meant to say that the quotient ring $$\Bbb Z[x] / (p \Bbb Z [x])$$ identifies with $$\Bbb F_p [x] := (\Bbb Z / p \Bbb Z)[x]$$: this is a ring of polynomials in one variable over a field, hence it is a PID. You can check that this identification holds by showing that the obvious map $$\Bbb Z[x] \rightarrow \Bbb F_p [x]$$ is surjective, and has kernel the ideal $$(p \Bbb Z [x])$$.

For your second question, we use the following fact:

Let $$A$$ be a ring and $$I$$ an ideal of $$A$$. Then the quotient map $$\pi:A \rightarrow A/I$$ induces a bijection between the sets $$\{\text{ideals of A containing I}\}$$ and $$\{\text{ideals of A/I}\}$$.

This correspondence works as you'd expect, you can try to show it by hand. In particular, an ideal $$J$$ of $$A$$ containing $$I$$ corresponds to $$\pi(J)$$, which is an ideal of $$A/I$$ because $$\pi$$ is surjective (exercice). In our context, $$A = \Bbb Z[x]$$, $$I = (2) = 2 \Bbb Z[x]$$, $$J = (2,x)$$, and you can see that $$\pi(J) = x \Bbb F_2[x] = (x)$$, because 2 is mapped to 0 by $$\pi$$.

The term proper ideal simply means that the ideal is not the whole ring. The authors added the term "maximal" as a further remark that the image is a maximal ideal (which are, as part of the definition, proper). As far as I am aware, "proper maximal" does not mean anything.

If $$p \neq 2$$, letting $$I = p \Bbb Z[x]$$ as above, you can see this time that $$\pi(2) \in \Bbb F_p[x]$$ is the constant $$\overline{2} \in \Bbb F_p$$, which is therefore an invertible element, i.e. a unit, of $$\Bbb F_p[x]$$ (its inverse is $$\overline{2}^{-1} \in \Bbb F_p$$, which exists because $$\Bbb F_p$$ is a field and $$\overline{2} \neq \overline{p}$$ is not zero). In such a case, the ideal $$\pi((2,x))$$ contains $$\overline{2}=\pi(2)$$ which is invertible by what we just argued, so by multiplying by its inverse, we see that the ideal $$\pi((2,x))$$ contains $$\overline{1}$$: it is then the whole ring $$\Bbb F_p[x]$$.

• $\Bbb{F}_p$ is only a field when $p\neq 2$ and $p$ is prime? That is the reason why $2\in \Bbb{F}_p[x]$ and has an inverse. Also, in the case where $p$ is prime, $\Bbb{F}_p$ is what is called a finite field?
– Seth
Commented Jul 15 at 11:26
• The ring $\Bbb Z/n\Bbb Z$ is indeed a field if and only if $n$ is a prime number, including 2 ($\Bbb Z / 2 \Bbb Z$ is a field, containing two elements). When this is the case, we usually denote $n$ by $p$, and yes, $\Bbb Z/p\Bbb Z$ is a field that is finite (so a finite field). In that case, to emphasize on the fact that it is a field (more than a ring), we write it as $\Bbb F_p$ (and also because it is quicker to write). Commented Jul 15 at 11:32
• Notice that the integer $2$ is mapped to a nonzero element of $\Bbb F_p$ except when $p=2$. So when $p \neq 2$, $2$ is invertible in $\Bbb F_p$. Commented Jul 15 at 11:33
• thank you so much for your detail write up. Is it okay if I trouble you also about this recent post about notation. There is a minor issue about the notation $j$ within the theorem reference in that post that I am not clear on. Thank you in advance.
– Seth
Commented Jul 15 at 11:35