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Suppose that $f(x) \in \Bbb Z_p[x]$ and is irreducible over $\Bbb Z_p$, where $p$ is prime. If $\deg f(x) = n$, prove that $\Bbb Z_p[x] / \langle f(x) \rangle$ is a field with $p^n$ elements.

I can see that $\Bbb Z_p[x] / \langle f(x) \rangle$ is a field because $\Bbb Z_p$ is a field and $f(x)$ is irreducible, but I can't figure a way to show that it has $p^n$ elements.

I know I can write any $g(x) + \langle f(x) \rangle$ as $q(x)f(x) + r(x) + \langle f(x) \rangle$ where $r(x)$ is the remainder upon division, but I can't seem to figure what the remainder would be when dividing arbitrary elements.

user26857
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Oliver G
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1 Answers1

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Regarding the $p^n$ elements in the field $\mathbb{Z}_p [x] / \langle f(x) \rangle$, note that any $q \in \mathbb{Z}_p [x] / \langle f(x) \rangle$ has the form $q(x) = \sum_{i = 0}^{n - 1} \alpha_i x^i$, where $\alpha_i \in \mathbb{Z}_p$ for any $0 \leq i \leq n -1$. From this it follows that there are $p^n$ different polynomials.

That $\mathbb{Z}_p [x] / \langle f(x) \rangle$ has $p^n$ elements also immediately follows from the fact that $\mathbb{F}_{p^n} \cong \mathbb{Z}_p [x] / \langle f(x) \rangle$, where $\mathbb{F}_{p^n}$ denotes the Galois field. Of course you should be aware with this isomorphism. Therefore, the first approach is more straightforward.

  • Why is any $q$ of that form? I can't see how you got that with the division algorithm. – Oliver G Sep 15 '16 at 13:58
  • Each element $q$ in the field is of the form $g + \langle f(x) \rangle$, which is in fact the remainder in the division of $g$ by $f$. Recall that $deg(q) < deg (f) = n$, so at most $deg(q) = n-1$. –  Sep 15 '16 at 14:07
  • Why is the degree of $q$ less than $f$? – Oliver G Sep 15 '16 at 14:28
  • The polynomial $q$ is the remainder in the division of $g$ by $f$, so therefore $deg(q) < deg(f)$. This fact should be stated in the division algorithm to which you referred, isn't it? –  Sep 15 '16 at 14:29
  • I'm still confused. You're saying that any element $q$ in $\Bbb Z_p[x]/\langle f(x) \rangle$ is of the form $q(x) = \sum_{i=0}^{n-1}=\alpha_ix^i$ which is the remainder by the division algorithm, but how do you get $q(x) = \sum_{i=0}^{n-1}=\alpha_ix^i$ using the division algorithm? I can't find a way to use the division algorithm to show that any element is of that form. I see that dividing $g(x)$ by $f(x)$ will give a remainder, but I don't understand how I can find the general remainder to be in that form. – Oliver G Sep 15 '16 at 14:33
  • @OliverG If $\deg g\geq\deg f$ you can keep subtracting multiples of $f$ to lower the degree. So eventually $\deg g<\deg f$. – Servaes Sep 15 '16 at 14:34
  • Let $g \in \mathbb{Z}_p [x]$. Then $g$ can be written in the form $g(x) = h(x) f(x) + q(x)$ by the division algorithm. The elements in $\mathbb{Z}_p [x] / \langle f(x) \rangle$ are of the form $g + \langle f(x) \rangle$, which are the remainders $q(x)$ of degree at most $n - 1$. Any polynomial $q$ of degree $n - 1$ is of the form I've stated in my answer. –  Sep 15 '16 at 14:38
  • But how are you getting the remainder to be that particular sum? How are you sure that there no other possibilities for the remainder? – Oliver G Sep 15 '16 at 14:40
  • @OliverG. I guess it is clear that any polynomial of degree $n$ is of the form $\sum_{i = 0}^n a_i x^i$, isn't it? This is the definition of a polynomial. The elements in $\mathbb{Z}_p [x] / \langle f(x) \rangle$ are all the remainders of $g$ by $f$, and thus all polynomials of degree $n - 1$. –  Sep 15 '16 at 14:48
  • I think I see it now. Because if you divide $g(x)$ by $f(x)$ where deg$g(x) >$ deg $f(x)$, you can always eliminate the first term, but the others may not be eliminated and therefore cause the set of remainder polynomials. And it's $p^n$ different remainders by counting principals, right? – Oliver G Sep 15 '16 at 15:00
  • Yes, exactly. For a quotient ring $R / I$, the elements are always the residue classes modulo $I$. Thus in your example, the residue classes are all $g \in \mathbb{Z}_p [x]$ modulo $\langle f(x) \rangle$, which is what you described. –  Sep 15 '16 at 15:05
  • I have no idea what the OP has understood from this answer (and I don't have time to read the above comments), but: 1. You should prove that there are exactly $p^n$ elements in the quotient ring. (Their form is not a proof for their number.) 2. Saying that the quotient ring is isomorphic to $\mathbb F_{p^n}$ has no meaning without proving before that the quotient ring is a field with $p^n$ elements. As a conclusion, this answer is a mess of claims without proof. (-1) – user26857 Sep 16 '16 at 03:08
  • @user26857. Thanks for giving at least some comments on your downvote. I disagree with your comments though. I do not claim to give a thorough proof. 1) the reason I mentioned the form of the elements is because it is clear from this that it is a polynomial of degree $< n$, and to speak with your own words, then showing that the number of elements in the quotient ring is $p^n$ is "easily enough". This was not a problem for the OP either. 2) it has meaning to know the isomorphism without knowing the above result as $\mathbb{F}_{p^n}$ clearly has $p^n$ elements. This is less trivial in the ring. –  Sep 16 '16 at 07:12