Question on proof regarding structure of finite fields

**AlexP** · September 20th 2011, 12:29 PM

I am reading the proof of the fact that as a group under multiplication, the set of nonzero elements of $\mbox{GF}(p^n)$ (the Galois field/finite field of order $p^n$ ) is isomorphic to $\mathbb{Z}_{p^n-1}$ and is therefore cyclic.

During the course of the proof, the author reminds the reader that the operation is componentwise addition. I do not see why this is true...we are looking at the multiplicative group of the field. Can someone explain? (I will add more detail if necessary).

**Drexel28** · September 20th 2011, 07:27 PM

Originally Posted by AlexP

I am reading the proof of the fact that as a group under multiplication, the set of nonzero elements of $\mbox{GF}(p^n)$ (the Galois field/finite field of order $p^n$ ) is isomorphic to $\mathbb{Z}_{p^n-1}$ and is therefore cyclic.

During the course of the proof, the author reminds the reader that the operation is componentwise addition. I do not see why this is true...we are looking at the multiplicative group of the field. Can someone explain? (I will add more detail if necessary).

This doesn't make sense, we need more context. What would he even mean by component wise addition? Perhaps thinking of $\mathbb{F}_{p^n}$ in its usual form as $\mathbb{Z}[x]/(p(x))$ for an appropriate $p(x)$ from where the 'addition componentwise' would make sense for the addition in $\mathbb{F}_{p^n}$ .

**AlexP** · September 21st 2011, 08:20 PM

True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

"To see that the multiplicative group $\mbox{GF}(p^n)^\times$ of nonzero elements of $\mbox{GF}(p^n)$ is cyclic, we first note...that it is isomorphic to a direct product of the form $\mathbb{Z}_{n_1} \otimes \mathbb{Z}_{n_2} \otimes \cdots \otimes \mathbb{Z}_{n_k}$ , where each $n_{i+1}$ divides $n_i$ . So, for any element $a=(a_1, a_2, ..., a_k)$ in this product, we have $a^{n_1} = (n_1a_1, n_1a_2, ..., n_1a_k)$ . (Remember, the operation is componentwise addition.)
Thus, the polynomial $x^{n_1}-1$ has $p^n-1$ zeros in $\mbox{GF}(p^n)$ . Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that $p^n-1 \le n_1$ . On the other hand, since $\mbox{GF}(p^n)^\times$ has a subgroup isomorphic to $\mathbb{Z}_{n_1}$ , we also have $n_1 \le p^n-1$ . It follows, then, that $\mbox{GF}(p^n)^\times$ is isomorphic to $\mathbb{Z}_{p^n-1}$ ."

**Drexel28** · September 21st 2011, 08:27 PM

Originally Posted by AlexP

True...the componentwise part doesn't make sense without the rest. Here's the whole proof. There's another part I don't understand so I'll have that part ready too."

"To see that the multiplicative group $\mbox{GF}(p^n)^\times$ of nonzero elements of $\mbox{GF}(p^n)$ is cyclic, we first note...that it is isomorphic to a direct product of the form $\mathbb{Z}_{n_1} \otimes \mathbb{Z}_{n_2} \otimes \cdots \otimes \mathbb{Z}_{n_k}$ , where each $n_{i+1}$ divides $n_i$ . So, for any element $a=(a_1, a_2, ..., a_k)$ in this product, we have $a^{n_1} = (n_1a_1, n_1a_2, ..., n_1a_k)$ . (Remember, the operation is componentwise addition.)
Thus, the polynomial $x^{n_1}-1$ has $p^n-1$ zeros in $\mbox{GF}(p^n)$ . Since the number of zeros of a polynomial over a field cannot exceed the degree of the polynomial, we know that $p^n-1 \le n_1$ . On the other hand, since $\mbox{GF}(p^n)^\times$ has a subgroup isomorphic to $\mathbb{Z}_{n_1}$ , we also have $n_1 \le p^n-1$ . It follows, then, that $\mbox{GF}(p^n)^\times$ is isomorphic to $\mathbb{Z}_{p^n-1}$ ."

First of all, don't use tensor for direct product since that has a separate meaning. What he is really saying is this, if $f:\mathbb{F}_{p^n}\to\mathbb{Z}_{n_1}\times\cdots \times\mathbb{Z}_{n_k}$ is an isomorphism we have that $x^{n_1}-1=f^{-1}(f(x^{n_1}-1))=f^{-1}(n_1f(x)-1)=f^{-1}(0)=1$ where I made use of the fact that $n_1f(x)-1=0$ since, if $f(x)=(a_1,\cdots,a_k)$ we have that $n_1(a_1,\cdots,a_k)=(na_1,\cdots,na_k)=(0,\cdots,0 )$ since $n_1\mid |\mathbb{Z}_{n_j}|$ for each $j\in[k]$ and so $n_1a_j=0$ by Lagrange's theorem. Make sense?

Also if you are having trouble understanding this I, at the risk of shamelessly self-promoting, suggest you see this post I wrote on my blog about this topic.

**AlexP** · September 22nd 2011, 02:31 PM

I understand that $n_1(a_1, ..., a_k)=(n_1a_1, ..., n_1a_k) = (0,...,0)$ , but with $x^{n_1}-1=\cdots=0$ I'm getting $x^{n_1}=2$ which does not make sense.

That notation was a mistake, no idea what I was thinking. I meant to use what the book uses, which is $\oplus$ .

**NonCommAlg** · September 22nd 2011, 03:24 PM

more generally, any finite subgroup of the multiplicative group of a field is cyclic.
this result, under some conditions, can be extended nicely to division rings. see here.

**Drexel28** · September 22nd 2011, 03:42 PM

Originally Posted by AlexP

I understand that $n_1(a_1, ..., a_k)=(n_1a_1, ..., n_1a_k) = (0,...,0)$ , but with $x^{n_1}-1=\cdots=0$ I'm getting $x^{n_1}=2$ which does not make sense.

That notation was a mistake, no idea what I was thinking. I meant to use what the book uses, which is $\oplus$ .

How exactly do you manage that? Perhaps if you told us how you are getting that result we can help find your mistake.

**AlexP** · September 23rd 2011, 09:05 AM

Wow...I have no idea where my brain has been lately. That's scary.

What you did say was $x^{n_1}-1= \cdots =1$ , and that's where I got that. I'm clearly not following something.

**Drexel28** · September 23rd 2011, 10:19 AM

Originally Posted by AlexP

Wow...I have no idea where my brain has been lately. That's scary.

What you did say was $x^{n_1}-1= \cdots =1$ , and that's where I got that. I'm clearly not following something.

I think that it's perhaps because I made an EGREGIOUS typo/mistake in my earlier post (what time of night was that)? Ok, so you're trying to show that $x^{n_1}=1$ but since $f$ is an isomorphism this is EQUIVALENT to showing that $f(x^{n_1})=f(1)=0$ (we know that $f(1)=0$ since $1$ is the identity of $\mathbb{F}_{p^n}$ and so cooresponds to the identity $0=(0,\cdots,0)$ of $\mathbb{Z}_{n_1}\times\cdots\times\cdots\times \mathbb{Z}_{n_m}$ ). But, since $f$ is a morphism we know that $f(x^{n_1})=n_1f(x)$ from there I think you got the reason why this is zero.

I'm sorry for the error, does that make more sense?

**AlexP** · September 23rd 2011, 12:16 PM

It's fine, everyone's entitled to some mistakes (I certainly made some as well). I do see it clearly now. Just one thing... You wrote our map as $f: \mathbb{F}_{p^n} \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}$ , but should it really be $f: \mathbb{F}_{p^n}^\times \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}$ (specifically from the multiplicative group of our field to the direct product)?

**Drexel28** · September 23rd 2011, 02:28 PM

Originally Posted by AlexP

It's fine, everyone's entitled to some mistakes (I certainly made some as well). I do see it clearly now. Just one thing... You wrote our map as $f: \mathbb{F}_{p^n} \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}$ , but should it really be $f: \mathbb{F}_{p^n}^\times \to \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_k}$ (specifically from the multiplicative group of our field to the direct product)?

Yes.

**AlexP** · September 23rd 2011, 09:07 PM

ok. Thank you very much for the help (and patience).

**Drexel28** · September 23rd 2011, 09:09 PM

Originally Posted by AlexP

ok. Thank you very much for the help (and patience).

Anytime

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