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!

  • $\begingroup$ How about $-1$? $\endgroup$ – Tobias Kildetoft Mar 28 '19 at 7:59
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Mar 28 '19 at 8:03
  • $\begingroup$ 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 $5$ $\endgroup$ – Maxime Ramzi Mar 28 '19 at 9:41
  • $\begingroup$ What is the source of this question? Are you sure it is well-posed? It seems to me that there are no such units... $\endgroup$ – 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 $0 \leqslant k \leqslant 4$.

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  • $\begingroup$ 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? $\endgroup$ – Riju Mar 29 '19 at 0:39
  • $\begingroup$ 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! $\endgroup$ – Riju Mar 29 '19 at 0:43
  • $\begingroup$ @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. $\endgroup$ – Alex Wertheim Mar 29 '19 at 1:34
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    $\begingroup$ 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 $\mathbb{Z}[G]$ $\endgroup$ – 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 $\mathbb{Z}\times \mathbb{Z}[\zeta_5]$.

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  • $\begingroup$ In terms of elements of the group ring what kind of element is a unit? Any explicit one? $\endgroup$ – Riju Mar 28 '19 at 8:31
  • $\begingroup$ $\zeta_5$ 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. $\endgroup$ – Seewoo Lee Mar 28 '19 at 8:34
  • $\begingroup$ 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. $\endgroup$ – Alex Wertheim Mar 28 '19 at 22:05
  • $\begingroup$ @AlexWertheim You are right, thank you for sharing. $\endgroup$ – Seewoo Lee Mar 28 '19 at 23:24
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    $\begingroup$ @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. $\endgroup$ – Alex Wertheim Mar 28 '19 at 23:35

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