# Multiplicative group modulo polynomials

When working over $\mathbb{Z}$, it is well known what the structure of the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^{\times}$ exactly is: If $n=p$ for a prime $p$, then $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is the $p-1$ cyclic group $C_{p-1}$. If $n=p^e$ for an odd prime $p$, then $(\mathbb{Z}/p^e\mathbb{Z})^{\times}$ is again a cyclic group $C_{\phi(p^e)}=C_{p^e-p^{e-1}}$. If $n=2^e$ for $e \geq 2$, then $(\mathbb{Z}/2^e\mathbb{Z})^{\times}$ is no longer a cyclic group, but a product of two cyclic groups. It is of the form $C_{2} \times C_{2^{e-2}}$. For a more general integer $n$, we can simply use the Chinese remainder theorem to obtain the precise structure of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ using the above results.

My question is:

Is there a similar result for the multiplicative group of a quotient ring of $\mathbb{F}_{p}[X]$ where $\mathbb{F}_{p}$ is the finite field of $p$ elements (where $p$ is a prime power)? That is, what is the structure of $$(\mathbb{F}_{p}[X]/\left<f(X)\right>)^{\times}$$ for a polynomial $f(X) \in \mathbb{F}_{p}[X]$?

If $f(X)$ is irreducible, then this group is again cyclic (since we know that the multiplicative group of a finite field is cyclic). But what about the more general case? Again, we can use Chinese remainder theorem to reduce the general question to the case of $f(X)$ being a power of an irreducible. So what do we know about the structure of $$(\mathbb{F}_{p}[X]/\left<f(X)^e\right>)^{\times}$$ for an irreducible $f(X) \in \mathbb{F}_{p}[X]$ and $e \geq 2$?

• en.wikipedia.org/wiki/Finitely_generated_abelian_group Oct 6 '16 at 21:01
• @Bharat Have you tried to find out the group of units for $\mathbb F_p[X]/(X^n)$? Oct 6 '16 at 21:18
• See here, Proposition 2.4. Oct 7 '16 at 14:30
• @user26857 I did try that, but for simple cases like $X^2$ and $X^3$. Did not succeed in general. Thanks for the link. Oct 7 '16 at 19:03

what do we know about the structure of $({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}$ for an irreducible $f(X)\in \mathbb F_p[X]$ and $e\geq 2$?

Assuming that $\deg(f) = n$, I claim that $({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}$ is isomorphic to

$$\mathbb Z_{p^n-1}\times \left( \mathbb Z_{p}^{n(e-2\lceil\frac{e}{p}\rceil+ \lceil\frac{e}{p^2}\rceil)}\times \mathbb Z_{p^2}^{n(\lceil\frac{e}{p}\rceil -2\lceil\frac{e}{p^2}\rceil+ \lceil\frac{e}{p^3}\rceil)}\times \mathbb Z_{p^3}^{n(\lceil\frac{e}{p^2}\rceil -2\lceil\frac{e}{p^3}\rceil+ \lceil\frac{e}{p^4}\rceil)}\times \cdots\right).$$

The product looks infinite, but the exponents on the factors eventually become zero. For example, the group $\left(\mathbb F_3[x]/\langle(x^2+1)^{13}\rangle\right)^{\times}$ is isomorphic to $\mathbb Z_8\times \left(\mathbb Z_3^{10}\times \mathbb Z_9^4\times \mathbb Z_{27}^2\right)$.

Let $R = ({\mathbb F}_p[X]/\langle f(X)^e\rangle)$ and $J = \langle f(X)\rangle$. Reduction modulo $J$ yields a homomorphism of $R^{\times}$ onto $(R/J)^{\times}$ with kernel $1+J$. The image $(R/J)^{\times}$ is isomorphic to $\mathbb F_{p^n}^{\times}\cong \mathbb Z_{p^n-1}$, which has order that is relatively prime to the order of the kernel $|1+J| = |J| = p^{n(e-1)}$. Thus $$R^{\times} \cong (R/J)^{\times}\times (1+J)\cong \mathbb Z_{p^n-1}\times (1+J).$$ It remains to determine the structure of the $p$-group $(1+J)$.

The structure of a finite abelian $p$-group $A$ is determined by the orders of the annihilators

$$A[p^k]:= \{a\in A\;|\;p^ka = 0\}.$$

These orders are not so hard to determine when $A = 1+J$. That is, for $\alpha\in J$, the element $1+\alpha\in 1+J$ is annihilated by $p^k$ exactly when $1=(1+\alpha)^{p^k} = 1+\alpha^{p^k}$, which happens exactly when $f^e$ divides $\alpha^{p^k}$, which happens exactly when $\alpha\in \langle f^{\lceil\frac{e}{p^k}\rceil}\rangle$. Thus

$$|A[p^k]| = |\langle f^{\lceil\frac{e}{p^k}\rceil}\rangle| = p^{n\left(e-\lceil\frac{e}{p^k}\rceil\right)}.$$

What remains is to determine the $m_i$'s in the expression

$$B = \mathbb Z_p^{m_1}\times \mathbb Z_p^{m_2}\times \mathbb Z_p^{m_3}\times \cdots$$

if this general abelian $p$-group has the same size annihilators as $A=1+J$; i.e., if $|B[p^k]| = |A[p^k]|$ for all $k$.

One computes that

$$\begin{array}{rl} |B[p]| &= p^{m_1+m_2+m_3+m_4+\cdots}\\ |B[p^2]| &= p^{m_1+2m_2+2m_3+2m_4+\cdots}\\ |B[p^3]| &= p^{m_1+2m_2+3m_3+3m_4+\cdots}\\ &\textrm{ETC.} \end{array}$$ Thus we need to solve the equations

$$\begin{array}{rl} m_1+m_2+m_3+m_4+\cdots&=n\left(e-\lceil\frac{e}{p}\rceil\right)\\ m_1+2m_2+2m_3+2m_4+\cdots&=n\left(e-\lceil\frac{e}{p^2}\rceil\right)\\ m_1+2m_2+3m_3+3m_4+\cdots&=n\left(e-\lceil\frac{e}{p^3}\rceil\right)\\ \textrm{ETC.}& \end{array}$$

The solution is $m_i = n\left( \lceil\frac{e}{p^{i-1}}\rceil-2\lceil\frac{e}{p^i}\rceil +\lceil\frac{e}{p^{i+1}}\rceil \right)$. Therefore

$$\begin{array}{rl} ({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}&=R^{\times}\\ &\cong (R/J)^{\times}\times (1+J)\\ &\cong \mathbb Z_{p^n-1}\times \left( \mathbb Z_{p}^{n(e-2\lceil\frac{e}{p}\rceil+ \lceil\frac{e}{p^2}\rceil)}\times \mathbb Z_{p^2}^{n(\lceil\frac{e}{p}\rceil -2\lceil\frac{e}{p^2}\rceil+ \lceil\frac{e}{p^3}\rceil)}\times \mathbb Z_{p^3}^{n(\lceil\frac{e}{p^2}\rceil -2\lceil\frac{e}{p^3}\rceil+ \lceil\frac{e}{p^4}\rceil)}\times \cdots\right). \end{array}$$

• Wow. Not as simple as I hoped for. Thanks. Wonder why this problem isn't widely discussed in elementary algebra textbooks when they deal with polynomial rings over fields. Oct 7 '16 at 11:42