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Let $\mathbb{F}_q$ be a finite field with $q$ elements and $GR(p^s,r)$ be a Galois ring with $p^{sr}$ elements and chacteristic $p^s$. I know that $$GR(p^s,r)/pGR(p^s,r)\cong \mathbb{F}_{p^r}.$$

A ring embedding is a injective ring homomorphism. How to find a ring embedding $\varphi$ from $\mathbb{F}_{p^r}$ to $GR(p^s,r)$?

If we write every elements $x\in \mathbb{F}_{p^r}$ as $(p_1,\cdots,p_r)$, then $\varphi(x)=(h(p_1),\cdots,h(p_r))$ should the be a ring embedding where $h$ is some kind of Hensel lift from $\mathbb{Z}_p$ to $\mathbb{Z}_{p^s}$. But I do not know how it works.

Zongxiang Yi
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1 Answers1

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Such an embedding cannot exist when $s>1$. $K=\Bbb{F}_{p^r}$ has characteristic $p$ so $p\cdot 1_K=0_K$. On the other hand the Galois ring $R=GR(p^s,r)$ has characteristic $p^s>p$. Meaning that $p\cdot 1_R\neq 0_R$.

But a ring homomorphism, such as an embedding $f:K\to R$, must satisfy $f(1_K)=1_R$ and $f(0_K)=0_R$. This is a contradiction, for $$ 0_R=f(0_K)=f(p\cdot 1_K)=p\cdot f(1_K)=p\cdot 1_R\neq0. $$

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Even in a (IMHO) bizarre universe where rings don't have multiplicative neutral elements preserved by homomorphisms, the answer is still the same. $f(1_K)$ cannot be an element of $pR$, for those are all nilpotent implying that $$f(1_K)=f(1_K^s)=f(1_K)^s=0_R$$ violating the wish that $f$ should be injective. If $f(1_K)\in R\setminus pR$, and the first argument kicks in.

Jyrki Lahtonen
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  • Pre-emptively posting this display of banality as a CW. I think we have a suitable umbrella duplicate somewhere. – Jyrki Lahtonen Jul 29 '19 at 10:05
  • Maybe I misunderstand something. For example, an "embedding" from $\mathbb{Z}3$ to $\mathbb{Z}{9}$ could be defined as $0\rightarrow0$, $1\rightarrow 3$, $2\rightarrow6$. In this case, $1\in \mathbb{Z}_3$ is not necessarily mapped to $1\in \mathbb{Z}_9$. What if the ring is without an identity ? – Zongxiang Yi Jul 30 '19 at 03:18
  • @zongxiangyi That is not an embedding of either rings or rngs. It does not preserve products. $1\cdot1$ does not map to $3\cdot3$. It does respect sums. Do you only want a map that is a homomorphism of additive groups? That is possible. – Jyrki Lahtonen Jul 30 '19 at 04:14