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While I have an understanding of some notions one learns in abstract algebra, that knowledge is extremely overshadowed by what I know in analysis. I opened up Jean-Pierre Serre's A Course in Arithmetic today having difficulty understanding most of it.

Let $K$ be an field. The image of $\mathbb{Z}$ in $K$ is an integral domain, hence isomorphic to $\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z}$, where $p$ is prime; its field of fractions is isomorphic to $\mathbb{Q}$ or to $\mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p$. In the first case, one says that $K$ is of characteristic zero; in the second case, that $K$ is of characteristic $p$.

So, I believe that the image of $\mathbb{Z}$ in $ K$ is an integral domain isomorphic to $\mathbb{Z}$ and its field of fractions isomorphic to $\mathbb{Q}$. However, I do not understand why the integral domain is isomorphic to the quotient groups in case two. The characteristic for the first case makes sense, but not for the second.

From my understanding $\mathbb{Z}/p\mathbb{Z}$ is finite with $p$ elements, but isn't the image of $\mathbb{Z}$ in $K$ always infinite thus not allowing the possibility of injectiveness to have an isomorphism? Obviously I am wrong somewhere.

Furthermore, I don't see where $p \choose k$ comes in for the first lemma

If char$(K)=p$, the map $\sigma: x \mapsto x^p$ is an isomorphism of $K$ onto one of its subfields $K^p$.

We have $\sigma(xy) = \sigma(x)\sigma(y)$. Moreover, the binomial coefficient $p \choose k$ is congruent to $0$ $(\bmod{p})$ if $0 < k < p$. From this it follows that $$\sigma(x+y) = \sigma(x) + \sigma(y)$$ hence $\sigma$ is a homomorphism. Furthermore, $\sigma$ is clear injective.

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    $\begingroup$ $\sigma(x+y)=(x+y)^p$, expand. All the middle terms are $0$ in $K$. $\endgroup$ – André Nicolas Jun 4 '15 at 7:07
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    $\begingroup$ No, the image of $\mathbb{Z}$ in $K$ is not always infinite; it's infinite iff $K$ has characteristic zero. For example, if $K = \mathbb{F}_p$ itself then the image is all of $K$, which has $p$ elements. $\endgroup$ – Qiaochu Yuan Jun 4 '15 at 7:08
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It's probably best if you look at a specific example to try to understand why the image of $\mathbb{Z}$ need not be infinite. Consider $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$. To make the distinction between the elements of $\mathbb{Z}$ and $\mathbb{F}_2$, I'll denote the latter elements with an overbar.

Then the relevant map $\varphi$ is the one given by assigning $\varphi(1)=\overline{1}$. Now we have $\varphi(2)=\varphi(1+1)=\varphi(1)+\varphi(1)=\overline{1}+\overline{1}=\overline{0}$, and thus we find that the kernel of $\varphi$ is exactly $2\mathbb{Z}$, and so we see that the characteristic of $\mathbb{F}_2$ is $2$.

The general idea is that there's no reason that $1+\cdots+1$ need be non-zero in general, and the least number of times (greater than zero) for which this happens is called the characteristic; if no such number exists, then we say the characteristic is zero.

As far as the binomial coefficient is concerned, the idea is that the binomial theorem $(x+y)^n = \sum_{k=0}^n{\binom{n}{k}x^ky^{n-k}}$ continues to hold, where $\binom{n}{k}$ is really the image of $\binom{n}{k}\in \mathbb{Z}$. $\binom{p}{k}$ has a factor of $p$ in it whenever $k\neq 0,p$, and so $\binom{p}{k}=0$ for $k\neq 0,p$ in $\mathbb{F}_p$.

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  1. The image of $\mathbb{Z}$ mentioned in the text is under the map $\varphi:\mathbb{Z}\to K$ that sends the integer $n$ to $\varphi(n)=n\cdot 1_K=\underbrace{1_K+1_K+\ldots+1_K}_{n\ \text{times}}$, if $n$ is non-negative and to minus that quantity if $n$ is negative. Now there are two possiblities, a) there is an integer $n$ such that $n.1_K=0$, or b) $n.1_K\neq 0$ for all $n\in \mathbb{Z}$.

In case a), it means the kernel contains $n\mathbb{Z}$. Since this kernel is a subgroup of $\mathbb{Z}$ is must be of the form $p\mathbb{Z}$ for some non-zero integer $p$ (not nec. prime at this point). The 1st isomorphism theorem tells you that $\mathbb{Z}/\ker{\varphi}=\mathbb{Z}/p\mathbb{Z}$ is isomorphic to a field (namely the subfield of $K$ generated by $1_K$), hence $p$ must be a prime number.

In case b), $\varphi$ is injective and you seems to have understand the situation here.

  1. Newton's binomial formula says

$(x+y)^p=\sum_{k=0}^n$ $p\choose k$ $x^k y^{(p-k)}$

Hence, mod p, only the terms $x^p$ and $y^p$ remain on the right hand side.

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  • $\begingroup$ I was still writing when Hayen's answer came out and I thought I would still post mine. I hope it is helpful. $\endgroup$ – Thibaut Dumont Jun 4 '15 at 7:19
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    $\begingroup$ For assurance, it does help. Thanks. $\endgroup$ – user240033 Jun 4 '15 at 7:26
  • $\begingroup$ Dumont, I have a question about notation. I took $K^p$ to mean the field with elements $x^p$. Is that true? $\endgroup$ – user240033 Jun 26 '15 at 8:21
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    $\begingroup$ Yes! $K^p=\{x^p\mid x\in K\}$. $\endgroup$ – Thibaut Dumont Jun 28 '15 at 10:29

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