Stillwell, Elements of Number Theory, exercise 12.2.1

Question

The exercise in question asks

Find the three nonzero elements of $\mathbb{Z}[\sqrt{-5}]/(2)$ and show that they are not zero divisors, hence $(2)$ is a prime ideal.

Screenshot of the exercise:

My attempt.

$(2)=\left\{2m+2n\sqrt{-5}: m,n\in\mathbb{Z}\right\}$ by definition of the principal ideal of the element $2$ of $\mathbb{Z}[\sqrt{-5}]$.

The three non-equal elements that produce three classes mod $(2)$ in the ring $\mathbb{Z}[\sqrt{-5}]$ are: $$ a=1,\,\, b=\sqrt{-5},\,\, c=1+\sqrt{-5}. $$ They are found as elements $m+n\sqrt{-5}$, where either $m$, $n$ or both are odd.

However, $c$ is a zero divisor: $$ c^2=(1+\sqrt{-5})^2=1+2\sqrt{-5}-5=2(-2+\sqrt{-5})\in (2). $$

Thus $\mathbb{Z}[\sqrt{-5}]/(2)$ is not an integral domain. But if $(2)$ were a prime ideal, then $\mathbb{Z}[\sqrt{-5}]/(2)$ would have been an integral domain. This means that $(2)$ is not a prime ideal, contrary to what the exercise asks.

Question. Where is my mistake?

Answer 1

Question: "Stillwell, Elements of NT, exercise 12.2.1: This means that (2) is not a prime ideal,"

Answer: The following holds:

$$\mathbb{Z}[\sqrt{−5}]/(2)≅\mathbb{Z}[x]/(2,x^2+5)≅\mathbb{Z}[x]/(2,x^2+1)≅A:=\mathbb{F}_2[x]/(x^2+1)$$

and the polynomial $f(x):=x^2+1$ is not irreducible in $\mathbb{F}_2[x]$. Hence the quotient ring $A$ is not an integral domain.

Note: A quotient $k[x]/(f(x))$ where $I:=(f(x))$ is the ideal generated by a polynomial $f(x)$, is an integral domain iff the polynomial $f(x)$ is irreducible. Hence in the above case since $f(x)$ is not irreducible it follows the ideal $f(x)$ is non-prime. In fact $(x+1)(x+1)=x^2+2x+1=x^2+1$ hence in the quotient $\mathbb{F}_2[x]/(x^2+1)$ it follows $z:=x+1$ is a zero divisor. The element $z$ is nilpotent.

The ring $A:=k[x]/(x^2+1)$ with $k:=\mathbb{F}_2$ is the following: It has a basis

$$A \cong k\{1, \overline{x} \}$$

as vector space over $k$ - the field with two elements $0,1$. Hence the 3 non-zero elements in $A$ are the elements $1,1+\overline{x}, \overline{x}$. Let $\overline{x}$ be the equivalence-class of $x$ in the quotient ring $A$. In the quotient ring $A$ it follows $x^2=-1=1$ hence it follows the elements $1, \overline{x}$ are a basis for $A$ as vector space over the field $k$.

Note: If you consider the ring $\mathbb{Z}[\sqrt{5}]/(2)$ you get the following calculation:

$$\mathbb{Z}[\sqrt{5}]/(2) \cong \mathbb{Z}[x]/(2,x^2-1) \cong k[x]/(x^2-1) \cong A:=k[x]/(x-1)(x+1)$$

where $k$ is the field with 2 elements. The ring $A$ is not an integral domain, hence $(2) \subseteq \mathbb{Z}[\sqrt{5}]$ is not a prime ideal.

If you search online you propably find a list of errata for the book - add this example to the list.

Answer 2

Just to expand on leoli1’s comment: this exercise is totally wrong, and whoever wrote it has overlooked some basic results in commutative algebra.

Since $\DeclareMathOperator{\Z}{\mathbb{Z}} \Z$ is a PID, then it has Krull dimension $1$, as its longest chains of prime ideals are of the form $(0) \subsetneq (p)$ where $p$ is any nonzero prime. Now $\Z[\sqrt{-5}]$ is an integral extension of $\Z$ (it is generated by $\sqrt{-5}$, a root of the monic polynomial $x^2 + 5$), and integral extensions preserve Krull dimension by the Cohen-Seidenberg theorems (see my answer here for a proof), so $\Z[\sqrt{-5}]$ also has Krull dimension $1$.

This makes the claim in the exercise impossible: all nonzero primes of $\Z[\sqrt{-5}]$ must be maximal. Otherwise, since every ideal is contained in a maximal ideal, a non-maximal prime ideal $\mathfrak{p}$ would lead to a chain of prime ideals $$ 0 \subsetneq \mathfrak{p} \subsetneq \mathfrak{m} $$ of length $2$, contradiction. Note that this argument holds in general for any integral extension of $\Z$, so in particular, any subring of the ring of integers of a number field.

To answer your comment on the other answer: the isomorphism $\frac{\mathbb{Z}[x]}{(x^2 + 5)} \overset{\sim}{\to} \mathbb{Z}[\sqrt{-5}]$ is given by $x \mapsto \sqrt{-5}$. The $4$ elements of $\mathbb{F}_2[x]/(x^2+1)$ can be represented by the polynomials of degree $< 2$, namely $0, 1, x, 1+x$, so the $4$ elements of $\Z[\sqrt{-5}]/(2)$ can be represented by $0, 1, \sqrt{-5}, 1 + \sqrt{-5}$.

Since $x^2 + 1 = (x+1)^2$ over $\mathbb{F}_2$, then $x+1$ is nilpotent in $\mathbb{F}_2[x]/(x^2+1)$. The same is true for the corresponding element $1 + \sqrt{-5} \in \Z[\sqrt{-5}]/(2)$: $$ (1 + \sqrt{-5})^2 = -4 + 2 \sqrt{-5} \equiv 0 \pmod{2} \, . $$