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I have a set $P=\{0,1,2*1,3*1,...,(p-1)*1\}$ where p is the smallest integer such that $p*1=0$ (here $0$ and $1 $ are additive and multiplicative identities).

I want to show that $P$ is a subfield of $F$, where $F$ a finite field with characteristic $p$ ($p$ is prime).

I'm having a hard time getting started on this question; so far I figured I'd start by showing that $P$ is closed under $+$ and $*$ but I'm getting stuck just doing that.

I was also wondering if there was an easier approach to solving this problem other than what I'm trying to do?

Edit: $P$ has the same operations as $F$

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    $\begingroup$ Can you describe exactly where you're getting stuck showing $P$ is closed under addition and multiplication? If you try writing down the sum of two elements of $P$ and showing it is in $P$, how far can you get? $\endgroup$ – Eric Wofsey Jan 17 '18 at 5:14
  • $\begingroup$ I started by taking arbitrary elements $a,b$ in $P$, so we have $a=x*1$ and $b=y*1$ for positive integers $x$ and $y$. So $a+b=x*1+y*1$, which is pretty much all I got... $\endgroup$ – one over pie Jan 17 '18 at 5:30
  • $\begingroup$ I started by taking arbitrary elements $a,b$ in $P$, so we have $a=x*1$ and $b=y*1$ for positive integers $x$ and $y$. So $a+b=x*1+y*1$, which is pretty much all I got... $\endgroup$ – one over pie Jan 17 '18 at 5:30
  • $\begingroup$ Keep in mind here that $*$ is not the same as multiplication ($\cdot$). $n*a$ for $n$ a natural number means $a$ added together $n$ times. Think about how you can simplify $x * 1 + y * 1$. $\endgroup$ – Ibrahim Tencer Jan 17 '18 at 5:49
  • $\begingroup$ The next step you have to take is using Distributive law. $\endgroup$ – P Vanchinathan Mar 14 at 3:50
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There is an easier approach. However, I'm not sure if it will be useful to you. It's easier depending on how much of the following you already know, and how much you have to prove from scratch.

For any ring, $R$, $R$ is an abelian group, so it is a $\newcommand{\ZZ}{\mathbb{Z}}\ZZ$-module, where if $n\in \ZZ$, and $r\in R$, $n\cdot r = \sum_{i=1}^n r$ if $n\ge 0$, and $n\cdot r = (-n)\cdot (-r)$ if $n < 0$.

In fact, because multiplication in a ring is distributive, $$n\cdot (rs)=\sum_{i=1}^n rs = r\sum_{i=1}^n s = r(n\cdot s)=\left(\sum_{i=1}^n r\right) s=(n\cdot r)s.$$ (For $n\ge 0$, for $n< 0$, use the positive $n$ case and the fact that $-(rs)=(-r)s=r(-s)$) Thus in fact $R$ is not just a $\ZZ$-module, it is a $\ZZ$-algebra.

Now if $R$ is unital, since $R$ is a $\ZZ$-algebra, the map $\phi:n\mapsto n\cdot 1$ is a ring homomorphism, $\phi:\ZZ\to R$. The kernel of this map is $p\ZZ$, where $p$ is the least positive integer such that $p\cdot 1=0$ (assuming such an integer exists). The image is then isomorphic to $\ZZ/p\ZZ$. Then if $R$ is a domain, $\ZZ/p\ZZ$ (being a subring of $R$) must be a domain. Hence $p$ must be prime. Thus $\ZZ/p\ZZ$ is a field. You can check that its image is of course $P=\{0,1,2\cdot 1,3\cdot 1,\ldots,(p-1)\cdot 1\}$.

To summarize: I've outlined the proofs of a bunch of basic facts above. You may already know these, and therefore not have to reprove them. In the shortest form of this argument, you can just say:

Let $k$ be your field, of characteristic $p>0$. Note that $P$ is the image of the natural map $\alpha:\ZZ\to k$, and that $\ker\alpha = p\ZZ$. Therefore the image of $\alpha$ is isomorphic to $\ZZ/p\ZZ$ by the first isomorphism theorem, which is a field. Hence $P$ is a subfield of $k$.

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  • $\begingroup$ I haven't learned about most of those things but thank you, it's some pretty cool stuff! $\endgroup$ – one over pie Jan 17 '18 at 6:13
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You were given that $P$ was just a set, so it doesn't even have an addition or a multiplication operation.

However, you can make your set $P$ into a commutative ring. The most useful way to do so is to identify it with the quotient ring $\mathbb{Z} / p \mathbb{Z}$.

Once you've done so, you can invoke the universal properties of free commutative rings and quotient rings:

  • For any ring $R$, there is a unique homomorphism $\mathbb{Z} \to R$
  • For any ring $R$, there is a bijective correspondence between
    • Homomorphisms $\mathbb{Z} / p \mathbb{Z} \to R$
    • Homomorphisms $f : \mathbb{Z} \to R$ with the property that $f(p) = 0$

And since $\mathbb{Z} / p \mathbb{Z}$ is a field, every homomorphism $\mathbb{Z} / p \mathbb{Z} \to R$ where $R$ is nonzero is automatically injective. (proof idea: what is the kernelof the homomorphism?)

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  • $\begingroup$ Sorry I should've specified that $P$ has the same operations as $F$. $\endgroup$ – one over pie Jan 17 '18 at 6:14
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A more simple solution that does not mention rings: Elements of $P$ can be written as sums of copies of $1$. You can prove closure under addition this way. You can also prove closure under $\cdot$ using distributivity:

\begin{align}(a \ast 1) \cdot (b \ast 1) &= \underbrace{(1 + 1 + \dots + 1)}_{a\text{ times}} \cdot \underbrace{(1 + 1 + \dots + 1)}_{b} \\ &= \underbrace{(1 \cdot 1) + (1 \cdot 1) + \dots + (1 \cdot 1)}_{ab} = (ab)\ast 1 \in P.\end{align}

Since $P \subseteq F$, the other field axioms also hold for $P$.

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