TUT ALG3 2

Problem 2

Question

Let $n\ge 1$ be an integer. Determine $U((\mathbb{Z}/n\mathbb{Z})[x])$.

Properties of nilpotents, units, and zero divisors in polynomial rings are developed in a more general setting in @conradNILPOTENTSUNITSZERO21 .

Let $\psi (n)$ be the squarefree part of $n\in \mathbb{N}$. Call $a\in \mathbb{Z}/n\mathbb{Z}$ nilpotent if $\psi (n)|a$. Call $g(x)\in \mathbb{Z}/n\mathbb{Z}[x]$ nilpotent if $g(x{)}^k=0$ for some $k\in \mathbb{N}$, called the order of $g$.

Lemma 300.1(Lemma).

The sum of two nilpotent polynomials is nilpotent.

Proof.

Let $g(x)$ and $h(x)$ be nilpotent polynomials of orders $\alpha$ and $\beta$. Then, $(g(x)+h(x){)}^{\alpha +\beta }=0$, since every term in its expansion is of the form $cg(x{)}^{i}h(x{)}^j$, where $i\ge \alpha$ or $j\ge \beta$.□

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Lemma 300.2(Lemma).

$g(x)={\sum }_{i\ge 0}{g}_i{x}^i$ is nilpotent iff every $g_i$ is nilpotent for $i\ge 0$.

Proof.

$(⟸)$ Let $s$ be the highest power of a prime factor in the prime factorization of $n$. It is clear that any product of $s$ or more nilpotent elements in $\mathbb{Z}/n\mathbb{Z}$ is $0$. Thus, $g(x{)}^s=0$, since every coefficient in $g(x{)}^s$ is a product of $s$ nilpotent elements.

$(⟹)$ If $h(x)$ is a nilpotent polynomial of degree $d$, it is clear that $h_0$ and $h_d$ must be nilpotent. Thus, any nilpotent polynomial of degree $1$ has nilpotent coefficients!

Now, we will proceed to prove the claim by induction. The claim is trivially true for $d=0$, is true for $d=1$ by the preceding observation. Next, assume that the claim is true for nilpotent polynomials of degree $d$. Let $h(x)=h_0+\dots h_{d+1}{x}^{d+1}$ be a nilpotent polynomial of degree $d+1$. Again, we know that $h_0$ and $h_{d+1}$ are nilpotent. Now, notice that

$$h_0+h_{1}x+\cdots +h_d{x}^d=h(x)+(-h_{d+1}{x}^{d+1}).$$

Since both $h(x)$ and $-h_{d+1}{x}^{d+1}$ are nilpotent, it follows from Lemma 1 that $h_0+h_{1}x+\cdots +h_d{x}^d$ is nilpotent. It follows from the induction hypothesis that $h_0,\dots ,h_d$ are nilpotent.□

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Now, I claim the following:

Claim 300.3(Claim).

$f(x)=\sum f_i{x}^i$ is a unit in $\mathbb{Z}/n\mathbb{Z}[x]$ $⟺$ $f_0\in U_n$ and $f_i$ is nilpotent for $i>0$.

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$(⟸)$ If $f_0\in U_n$ and $f_i$ is nilpotent for $i>0$, $f$ can be expressed as

$$f_0(1+f_0^{-1}\stackrel{g(x)\equiv }{\overbrace{(f_{1}x+f_2{x}^2+\dots f_k{x}^k)}}).$$E1

Note that by Lemma 2, $g(x)$ is nilpotent.

If we attempt to construct an inverse $h(x)$ of $1+f_0^{-1}g(x)$, we will observe $h(x)$ is forced to be¹

$$1-f_0^{-1}g(x)+(f_0^{-1}{)}^{2}g(x{)}^2+\cdots =\sum (-f_0^{-1}{)}^{i}g(x{)}^i.$$E2

It is clear that $h(x)\in \mathbb{Z}/n\mathbb{Z}[x]$ only if $g(x{)}^i=0$ for all $i>l$ for some integer $l$. Since $g(x)$ is nilpotent, this condition is satisfied. Thus, $f(x)$ is a unit, with inverse

$$f_0^{-1}\sum _{i=0}^l(-f_0^{-1}{)}^{i}g(x{)}^i.$$

$(⟹)$ $f_0\in U_n$ is immediate. We again express $f(x)$ as in (E1). Since $f_0$ is invertible and $f(x)$ is invertible, so is $1+f_0^{-1}g(x)$, the inverse of which must² then be given by (E2), because any inverse would need to satisfy the recursive relation from multiplying out, which formally yields the series. Since the inverse is a (finite-degree) polynomial, the series must terminate, implying $g(x)$ must be nilpotent, which in turn implies $f_i$ is nilpotent for $i>0$ by Lemma 2.

Neat result as a by-product

The sum of a unit element and a nilpotent element is a unit when they commute.

Proof.

If$u$ is a unit with inverse $v$, $a$ is nilpotent, and $au=ua$, we have $u+a=u(1+va)$. Note that $au=ua$ implies $va=av$, which implies $va$ is nilpotent. Thus, $1+va$ is invertible by the same argument we used in the previous proof.□

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Problem 8

Question

Determine all prime ideals in $\mathbb{Z}[x]$.

$f(x)\in \mathbb{Z}[x]$ is said to be irreducible modulo prime $p$ if it cannot be expressed as $r(x)+pg(x)$, where $r(x)$ is reducible. $f(x)\in \mathbb{Z}[x]$ is irreducible modulo $p$ iff $\overline{f(x)}\in {\mathbb{Z}}_p[x]$ is irreducible.

Lemma 300.4(Lemma).

If $q(x)=r(x)+pg(x)$ where $r(x)$ is reducible and $\text{deg}r(x)\ge \text{deg}q(x)$, there exist $r^{\prime }(x)$ and $g^{\prime }(x)$, where $r^{\prime }(x)$ is reducible and $\text{deg}{r}^{\prime }(x)=\text{deg}q(x)$, such that $q(x)=r^{\prime }(x)+p{g}^{\prime }(x)$.

Proof.

Assume$\text{deg}r(x)>\text{deg}q(x)$. Let $r(x)=\left({\sum }_{i=0}^n{a}_i{x}^i\right)\left({\sum }_{i=0}^m{b}_i{x}^i\right)$. Either $p|a_n$ or $p|b_m$. WLOG, assume the former. Let $a_n=kp$. Then, we have

$$q(x)=\underset{r^{\prime }(x)}{\underbrace{\left(\sum _{i=0}^{n-1}{a}_i{x}^i\right)\left(\sum _{i=0}^m{b}_i{x}^i\right)}}+p{g}^{\prime }(x).$$

This can be repeated until the degree of $r^{\prime }(x)$ is equal to the degree of $q(x)$.□

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Claim 300.5(Claim).

$I$ is a prime ideal in $\mathbb{Z}[x]$ iff one of the following holds:

1. $I=(0)$.
2. $I=(p)$ for some prime $p$.
3. $I=(q(x))$ for some irreducible $q(x)\in \mathbb{Z}[x]$.
4. $I=(p,f(x))$ for some prime $p$, where $f(x)$ is irreducible modulo $p$.

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$(⟹)$

Case 1: $I$ has constant polynomials.

Let $c\in I$ be a constant polynomial. At least one prime factor $p$ of $c$ must be in $I$, since $I$ is a prime ideal. If $q\ne p$ is prime and $q\in I$, $1\in I$ and $I=\mathbb{Z}[x]$. Thus, $p$ is the only prime in $I$, and all constants of $I$ are of the form $kp$.

Let $S=\{q(x)\in I:p\nmid q(x)\}$. If $S=\varnothing$, then $I=(p)$. Else, let $f(x)$ a minimum degree polynomial in $S$. Whatever be the value of the leading coefficient of $f(x)$, note it it cannot be divisible by $p$, and hence, using Bezout’s lemma, the leading coefficient of $f(x)$ can be reduced to $1$. Call this monic polynomial $f^{\ast }(x)$. Clearly, $f^{\ast }(x)$ is unique modulo $p$. Let the degree of $f^{\ast }(x)$ be $d$.

Note that $f^{\ast }(x)$ is not reducible modulo $p$; Indeed, if that were true, $f^{\ast }(x)$ could be expressed as $r(x)+pa(x)$ where $r(x)$ is reducible, and it would follow from Lemma 4 that there exists $r^{\prime }(x)=r_1^{\prime }(x)r_2^{\prime }(x)$ such that $\text{deg}{r}^{\prime }(x)=\text{deg}{f}^{\ast }(x)$ and

$$f^{\ast }(x)-p{a}^{\prime }(x)=r^{\prime }(x)\in I.$$

Since $I$ is a prime ideal, either $r_1^{\prime }(x)\in I$ or $r_2^{\prime }(x)\in I$ (neither of which are divisible by $p$), contradicting the minimality of the degree of $f^{\ast }(x)$.

We now claim that $I=(p,f^{\ast }(x))$. Let $u(x)={\sum }_{i=0}^{d^{\prime }}{u}_i{x}^i\in I$ have degree $d^{\prime }$. If $d^{\prime }<d$, $u(x)$ is divisible by $p$ by construction of $f^{\ast }(x)$. Let $d^{\prime }=d$.

$$\begin{array}{r}u(x)-u_{d^{\prime }}{f}^{\ast }(x)=v(x)\in I.\end{array}$$

Since $\text{deg}v(x)<d$, it follows that $v(x)=p{v}^{\prime }(x)$. Thus,

$$u(x)=f^{\ast }(x)u_{d^{\prime }}+p{v}^{\prime }(x).$$

For $d^{\prime }>d$,

$$\begin{array}{rl}& u(x)-u_{d^{\prime }}{x}^{d^{\prime }-d}{f}^{\ast }(x)=v(x)\in I\end{array}$$

By IH, $v(x)\in (p,f^{\ast }(x))$, so it follows that $u(x)\in (p,f^{\ast }(x))$.

Case 2: All polynomials in $I$ have degree at least $1$.

Warning

The following reasoning is flawed; see this instead.

Let $q(x)\in I$ be the minimum degree minimum leading coefficient polynomial. This clearly has to be unique. It has to be irreducible, since if $q(x)=r_1(x)r_2(x)$, either $r_1(x)\in I$ or $r_2(x)\in I$, a contradiction. All $u(x)\in I$ being divisible by $q(x)$ is also forced. Thus, $I=(q(x))$.

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$(⟸)$

If $I=(p)$ or $I=(q(x))$ for irreducible $q(x)$, $I$ is clearly a prime ideal. Assume $I=(p,q(x))$, where $q(x)$ is not reducible modulo $p$. Note that

$$\frac{\mathbb{Z}[x]}{(p,q(x))}\cong \frac{{\mathbb{Z}}_p[x]}{(\overline{q(x)})}.$$

Since $q(x)$ is not reducible modulo $p$, $\overline{q(x)}$ is not reducible in ${\mathbb{Z}}_p[x]$.

Claim 300.6(Claim).

$(\overline{q(x)})$ is a maximal ideal in ${\mathbb{Z}}_p[x]$.

Proof.

Let$I\subseteq {\mathbb{Z}}_p[x]$ be an ideal such that $(\overline{q(x)})$ is a proper subset of $I$. Let $\overline{g(x)}\in I\setminus (\overline{q(x)})$. Note that $\overline{g(x)}$ cannot be a multiple of $\overline{q(x)}$. If $\overline{g(x)}$ is a constant, that would imply $1\in I$ (${\mathbb{Z}}_p$ is a field!) and hence $I={\mathbb{Z}}_p$.

Assume $\text{deg}\overline{g(x)}\ge 1$. Let $\overline{f(x)}$ be the monic polynomial of smallest degree in $(\overline{q(x)},\overline{g(x)})$. Any $\overline{u(x)}\in (\overline{q(x)},\overline{g(x)})$ can be expressed as $\overline{u(x)}=\overline{f(x)}\cdot \overline{a(x)}+\overline{r(x)}$, where $\overline{r(x)}$ has smaller degree than $\overline{f(x)}$. However, $\overline{r(x)}=\overline{u(x)}-\overline{f(x)}\cdot \overline{a(x)}\in I$, forcing $\overline{r(x)}=0$. Thus, $(\overline{q(x)},\overline{g(x)})=(\overline{f(x)})$. This implies that $\overline{f(x)}$ is a factor of $\overline{q(x)}$, which forces $\overline{f(x)}=1$, and hence $I={\mathbb{Z}}_p$.□

Thus, ${\mathbb{Z}}_p[x]/(\overline{q(x)})$ (and thus $\mathbb{Z}[x]/(p,q(x))$) is a field, and in particular an integral domain. It follows that $(p,q(x))$ is a prime ideal in $\mathbb{Z}[x]$.

Footnotes

1. Ok, I haven’t been able to prove that $h(x)$ is forced to be (E2), but the argument here still works: Since $g(x)$ is nilpotent, (E2) must terminate, and is clearly an inverse of $1+f_0^{-1}g(x)$. ↩

2. This is where a problem arises. I absolutely NEED $h(x)$ to be (E2) for the argument to work. Refer @conradNILPOTENTSUNITSZERO21 Theorem 2.2 for a valid proof. ↩