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I need to prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element, but I'm not really sure how to begin.

I know that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ consists of all formal sums $w_{1}[z]_{1} + w_{2}[z]_{2} + \cdots + w_{k}[z]_{k}$ where the $w_{i}$ are integers, and the $[z]_{i}$ are elements of $\mathbb{Z}_{n}$. Now, my first question is, should these $[z]_{i}$ actually just be $\{1, 2, \cdots \, n\}$? I think being able to properly visualize what elements of this group ring look like might be rather helpful in figuring out how to do this problem.

Next, I know that $\mathbb{Z}_{n}$ itself can be generated by the single element $[1]_{n}$ and $\mathbb{Z}$ by $1$. How to extend this to a group ring is escaping me, however.

Any kind of push in the right direction would be welcome at this point by someone patient and willing to answer lots and lots of follow-up questions on my journey to understanding this completely.

Thank you ahead of time!

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    Downvoter, care to comment? – 2017-02-27
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    Why would you write $[z]_i$? Do you mean $[z_i]$ for $z_i\in\Bbb Z_n$? – 2017-02-27
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    @arctictern yes, that is what I mean. – 2017-02-27
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    Writing $x=[1]$, then $[2]=x^2$, $[3]=x^3$, etc. (with $1=[0]=[n]=x^n$). So every element looks like $a_0+a_1x+\cdots+a_{n-1}x^{n-1}$, and multiplication is as-usual for polynomials except that exponents of $x$ "wrap" back around as integers mod $n$. What would you say "generated by a single element" means? Since you're trying to prove $\Bbb Z[\Bbb Z_n]$ is generated by a single element, you should explicitly state what that means. – 2017-02-27
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    @arctictern by generated by a single element, I think what I mean is that there is one element $a_{0}+a_{1}x+\cdots + a_{n-1}x^{n-1}$ we can obtain all other elements in the group ring from. How we can do this, I'll admit I am not sure. I guess by whatever the operations in the group ring are (and what that is I am not entirely certain). – 2017-02-27
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    Take the element $x=[1]$. Every element of $\Bbb Z[\Bbb Z_n]$ can be created using just $x$, so it generates the ring. – 2017-02-27
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    @ALannister Since you know that the group is generated by a single element, reviewing the definition of addition and multiplication in the group ring may make the answer apparent. – 2017-02-27
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    @Qudit I can't seem to find that in my notes anywhere - I'll check the Wikipedia page on group rings. – 2017-02-27
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    @arctictern formally how would I express that? – 2017-02-27

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Let $C_n$ denote the cyclic group of order $n$, with generator $a$. The group ring $\mathbb{Z}[C_n]$ is precisely the ring $$R := \mathbb{Z}[x]/(x^n-1)$$ You should think of this presentation as saying: It is the ring which contains $\mathbb{Z}$ and contains "$a$", with the property that $a^n = 1$, ie $a^n-1=0$. This gives rise to the relation $x^n-1$. It's easy to check that both $R$ and $\mathbb{Z}[C_n]$ as $\mathbb{Z}$-modules are free of the same rank, so it's reasonable to think that they're isomorphic.

To prove that they are isomorphic, it suffices to define a ring homomorphism from one to the other, and to check that it is an isomorphism. The easiest way to do this would be to define a homomorphism from $\mathbb{Z}[x]\rightarrow \mathbb{Z}[C_n]$ which sends $x\mapsto a$. It's a homomorphism by the universal property of polynomial rings. Then it suffices to check that the kernel is precisely the ideal $x^n-1$. I haven't done this, but I think it should be reasonably straightforward.

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    that would be a great way to do it, but we haven't gotten to ring homomorphisms or ideals yet, so I don't think I can use those things. Is there a way to approach this problem without resorting to this method? – 2017-02-27
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    @ALannister I'm not really sure what the point of talking about group rings is before even learning about ideals. But in any case, it's pretty clear from the definition that the group ring is generated by "$a$" - ie, every element of the group ring is a formal sum of the form $c_0 + c_1a + c_2a^2 + \cdots + c_{n-1}a^{n-1}$, where $c_i\in\mathbb{Z}$. – 2017-02-27