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

Question

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!

Answer 1

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.

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

Answer 2

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 $0 \leqslant k \leqslant 4$.

Answer 3

I think solution given by Alex Wertheim is partially correct. The unit of $\mathbb{Z}[\zeta_5]$ $(1+\zeta_5)^2 = \zeta_5^2 + 2\zeta_5 + 1$ can be re-written as $\zeta_5+(\zeta_5^2 + \zeta_5 + 1) = \zeta_5-\zeta_5^3-\zeta_5^4$ and this has augmentation $-1$. $$\varphi(1.g-1.g^3-1.g^4)=(-1,(1+\zeta_5)^2)$$ So $(1+\zeta_5)^2$ is also an unit of $\mathbb{Z}[\zeta_5]$ with augmentation $-1$ and we get a non trivial unit of $\mathbb{Z}[G]$. This implies $(1+\zeta_5)^{2n}$ also has augmentation either $1$ or $-1$ because $\varphi$ is a ring homomorphism $$\varphi((1.g-1.g^3-1.g^4)^n)=(\varphi(1.g-1.g^3-1.g^4))^n=(-1,(1+\zeta_5)^2)^n = ((-1)^n, (1+\zeta_5)^{2n})$$ Also $\varphi(\frac{4}{5}+\frac{4}{5}.g-\frac{1}{5}.g^2-\frac{1}{5}.g^3-\frac{1}{5}.g^4)=(1,1+\zeta_5)$ by using the fact that $\zeta_{5}^{4}+\zeta_{5}^{3}+\zeta_{5}^{2}+\zeta_{5}+1 = 0$. Due to the isomorphism $\varphi$, this pre-image of $(1,1+\zeta_5)$ is unique. Similarly pre-image of $(-1,1+\zeta_5)$ also lies in $\mathbb{Q}[G]$ but not in $\mathbb{Z}[G]$.

As a result the group of units is $<-1.g, 1.g, 1.g-1.g^3-1.g^4>$ which is isomorphic to $\mathbb{Z_2} \times \mathbb{Z_5} \times \mathbb{Z}$

Answer 4

If $C_5=\langle g:g^5=1\rangle$ then in the group ring ${\Bbb Z}C_5$ holds $$(g^2+g^3-1)\cdot(g+g^4-1)=g^3+g-g^2+g^4+g^2-g^3-g-g^4+1=1.$$

Zbigniew Marciniak