# Obtain an unit in the integral group ring $\mathbb{Z}[\mathbb{Z}/5\mathbb{Z}]$

Obtain a nontrivial unit of the integral group ring $$\mathbb{Z}[\mathbb{Z}/5\mathbb{Z}]$$. By nontrivial I mean not a group element.

Since brute force doesn't seem possible here, I tried to guess what should be a unit, other than the group elements. But I have failed in this attempt. I would appreciate any kind of help.

Thank you!

• How about $-1$? – Tobias Kildetoft Mar 28 '19 at 7:59
• Try the analogues of cyclotomic units. If $c$ is a generator of the cyclic group of order five, then so is $c^k$ for all $k=2,3,4$. By the familiar formulas for geometric sums $1-c^k$ is then a factor of $1-c^\ell$ for all $k,\ell\in\{1,2,3,4\}$. Therefore $(1-c^k)/1-c^\ell)$ is a unit. Seewoo Lee's answer (+1) is about the same idea. – Jyrki Lahtonen Mar 28 '19 at 8:03
• Let me add a little note : if you reduce mod $5$ to get $\mathbb{F}_5[C_5]$, you get a local ring with maximal ideal the augmentation ideal. Hence its units are precisely the non elements of the augmentation ideal. Thus, if you're a unit in $\mathbb{Z}[C_5]$, your reduction mod $5$ is as well, and thus is not in the augmentation ideal : the units have augmentation that is not zero mod $5$ – Maxime Ramzi Mar 28 '19 at 9:41
• What is the source of this question? Are you sure it is well-posed? It seems to me that there are no such units... – Alex Wertheim Mar 28 '19 at 22:07

If I'm not mistaken, the answer by Seewoo Lee is not correct, I'm afraid. Putting $$G = \mathbb{Z}/5\mathbb{Z} =: \langle g \rangle$$, the group ring $$\mathbb{Z}[G]$$ is not isomorphic to $$\mathbb{Z}[\zeta_{5}]$$; it is isomorphic to $$\mathbb{Z}[X]/\langle X^{5}-1 \rangle$$, where the isomorphism is induced by the unique morphism of $$\mathbb{Z}$$-algebras $$\mathbb{Z}[X] \to \mathbb{Z}[G]$$ which sends $$X$$ to $$g$$. In fact, you can see that $$\mathbb{Z}[G]$$ has zero divisors, as illustrated in Max's comment to Thomas' answer.

Earlier, I had made a comment that $$\mathbb{Z}[X]/\langle X^{5}-1 \rangle \cong \mathbb{Z} \times \mathbb{Z}[\zeta_{5}]$$ by the Chinese remainder theorem, but this is not correct either, since the ideals $$\langle X-1 \rangle, \langle X^{4}+X^{3}+X^{2}+X+1 \rangle$$ in $$\mathbb{Z}[X]$$ are not comaximal. The matter is more subtle, but there is a way to at least produce units, as suggested by this wonderful answer of Dustan Levenstein's.

Over $$\mathbb{Q}$$, the ideals $$\langle X-1 \rangle, \langle X^{4}+X^{3}+X^{2}+X+1 \rangle$$ are comaximal, and so there is an isomorphism $$\varphi \colon \mathbb{Q}[G] \xrightarrow{~\sim~} \mathbb{Q}[X]/\langle X^{5}-1\rangle \xrightarrow{~\sim~} \mathbb{Q} \times \mathbb{Q}[\zeta_{5}]$$ defined by

$$\varphi(a_{0} + a_{1}g + a_{2}g^{2} + a_{3}g^{3}+a_{4}g^{4}) = \left(\sum_{i=0}^{4} a_{i}, ~\sum_{i=0}^{4} a_{i}\zeta_{5}^{i} \right)$$

This is the augmentation map on $$\mathbb{Q}[G]$$ in the first component. Composing $$\varphi$$ with the inclusion $$\mathbb{Z}[G] \hookrightarrow \mathbb{Q}[G]$$ gives us an injective morphism of rings which takes image in $$\mathbb{Z} \times \mathbb{Z}[\zeta_{5}] \subset \mathbb{Q} \times \mathbb{Q}[\zeta_{5}]$$. Hence, the restriction of $$\varphi$$ to $$\mathbb{Z}[G]$$ gives an injective morphism $$\mathbb{Z}[G] \to \mathbb{Z} \times \mathbb{Z}[\zeta_{5}]$$, so it suffices to find a unit of $$\mathbb{Z} \times \mathbb{Z}[\zeta_{5}]$$ which is in the image of $$\varphi$$. This amounts to finding an element of $$\mathbb{Z}[\zeta_{5}]$$ which is a unit and also has augmentation $$\pm 1$$.

Unfortunately, if I am not mistaken, there are no such elements other than $$\pm \zeta_{5}^{i}$$ for $$0 \leqslant i \leqslant 4$$, which says that the units of $$\mathbb{Z}[G]$$ are precisely the group elements, up to sign. (If someone sees an error here, please correct me!)

Indeed, it is asserted in Chapter 1, Section 7, Exercise 4 of Neukirch's "Algebraic Number Theory" that the units of $$\mathbb{Z}[\zeta_{5}]$$ are precisely:

$$\mathbb{Z}[\zeta_{5}]^{\times} = \{\pm \zeta_{5}^{k} (1+\zeta_{5})^{n} \mid 0 \leqslant k < 5, n \in \mathbb{Z}\}$$

Here, we note that $$(1+\zeta_{5})^{-1} = -\zeta_{5} - \zeta_{5}^{3}$$, since $$\zeta_{5}^{4}+\zeta_{5}^{3}+\zeta_{5}^{2}+\zeta_{5}+1 = 0$$. Since $$\zeta_{5}^{k}$$ has augmentation $$1$$ for any $$0 \leqslant k \leqslant 4$$, and $$(1+\zeta_{5})^{n}$$ has (up to sign) augmentation $$2^{|n|}$$ for any $$n \in \mathbb{Z}$$, it follows that the only units of $$\mathbb{Z}[\zeta_{5}]$$ with augmentation $$\pm 1$$ are $$\pm \zeta_{5}^{k}$$ for $$0 \leqslant k \leqslant 4$$.

• You are basically saying that the group elements are the only elements in the above$\mathbb{Z}[G]$, which are units, ofcourse upto sign. Now maybe I am having a basic confusion about the definition of a group ring. How does one sees a group elements in $\mathbb{Z}[G]$? I see it as follows: If $g\in G$, then the group element in $\mathbb{Z}$ is just $1.g$. Now, my question is that the following elements in the aforementioned group ring: $1.\bar{0}+1.\bar{1}$, is it a element of the group? I think that it is not same as $1.\bar{1}$. So maybe I am not getting the definition properly? – Riju Mar 29 '19 at 0:39
• As to your question about the source of the question, I was asked this question by a friend, so I really don't know about the source. I didn't know that one can find the structure of the group of units, by techniques used above, so I was tryng brute force method to find out a unit! – Riju Mar 29 '19 at 0:43
• @Riju: $\mathbb{Z}[G]$ is free as a $\mathbb{Z}$-module, with basis given by the elements of $G$. Thus, as you say: when one says a group element of $\mathbb{Z}[G]$, one means $1 \cdot g$ for $g \in G$. And indeed, you are right as well that $1 \cdot e + 1 \cdot g$ and $1 \cdot g$ are distinct elements. – Alex Wertheim Mar 29 '19 at 1:34
• I would have made the same mistake, thank you for pointing out that comaximal ideals in $\mathbb{Q}[G]$ aren't necessarily comaximal when "naturally restricted" to $\mathbb{Z}[G]$ – Maxime Ramzi Mar 29 '19 at 10:06

Hint: the group ring is isomorphic to $$\mathbb{Z}[\zeta_{5}]$$, where $$\zeta_5$$ is a 5th root of unity, which is a ring of integer $$\mathbb{Q}(\zeta_{5})$$. In fact, we can even compute the unit group - see Exercise 4 in Chapter 1.7 of Neukirch, Algebraic number theory. Note that $$1+\zeta_5$$ is a unit of infinite order.

Edit: As Alex Wertheim said, $$\mathbb{Z}[\mathbb{Z}/5\mathbb{Z}]$$ is NOT isomorphic to $$\mathbb{Z}[\zeta_5]$$, but isomorphic to $$\mathbb{Z}\times \mathbb{Z}[\zeta_5]$$.

• In terms of elements of the group ring what kind of element is a unit? Any explicit one? – Riju Mar 28 '19 at 8:31
• $\zeta_5$ corresponds to the element $1$ in $\mathbb{Z}/5\mathbb{Z}$, so we can write it as $u = 1\cdot (0) + 1\cdot (1)$, where $(0), (1)$ are elements in $\mathbb{Z}/5\mathbb{Z}$ and $u$ is a formal sum of these two elements. – Seewoo Lee Mar 28 '19 at 8:34
• If I am not mistaken, this answer is not correct. (The major error is that the group ring is not isomorphic to $\mathbb{Z}[\zeta_{5}]$, it is isomorphic to $\mathbb{Z}[X]/\langle X^{5}-1 \rangle$). Please see my answer for more details. – Alex Wertheim Mar 28 '19 at 22:05
• @AlexWertheim You are right, thank you for sharing. – Seewoo Lee Mar 28 '19 at 23:24
• @SeewooLee: no problem, but $\mathbb{Z}[\mathbb{Z}/5\mathbb{Z}]$ is not isomorphic to $\mathbb{Z} \times \mathbb{Z}[\zeta_{5}]$ either. (I made this mistake earlier as well.) The factors of $X^{5}-1$ do not generate comaximal ideals in $\mathbb{Z}[X]$. The situation is more complicated, and is detailed in my answer. – Alex Wertheim Mar 28 '19 at 23:35