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I do not understand what sets like these are. I know what something like $\mathbb{Z}_7$ is. It is the ring of integers modulo 7 so it is equal to ${0,1,2,3,4,5,6}$. But what is $\mathbb{F}_7[X]$ equal to. I don't understand. I have spent ages searching on the internet but can't find anything on it. All I know is that it has 7 elements but I just want to see a clear definition of it.

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    $\begingroup$ $\mathbb{F}_{7}[x]$ does not have $7$ elements. In fact, it has infinitely many elements. It is the set of all polynomials in $X$ with coefficients in $\mathbb{F}_{7}$. $\endgroup$ – Alex Wertheim Nov 17 '14 at 0:37
  • $\begingroup$ possible duplicate of ideals question with irreducibility $\endgroup$ – Ivo Terek Nov 17 '14 at 0:39
  • $\begingroup$ OK then what exactly is $\mathbb{F}_7$?? $\endgroup$ – snowman Nov 17 '14 at 0:40
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Well, you know what the ring $\mathbb{Z}_{n}$ is for any integer $n$. It turns out that if $n$ is a prime number (like $2$, $3$, $5$, $7$, etc.) (when $n$ is prime, it is usually replaced with the letter $p$) then $\mathbb{Z}_{p}$ is actually a field! It's not just a ring, but it also has a multiplicative identity (if your definition of ring doesn't already come with this assumption), no zero divisors, and every nonzero element has a multiplicative inverse.

Usually, people write $\mathbb{F}_{p}$ instead of $\mathbb{Z}_{p}$ since this is a field. Somehow, calling it $\mathbb{F}_{p}$ reminds us that it is the field of order $p$.

Now, what happens when we "adjoin $X$" to this field? That is, what is $\mathbb{F}_{p}[X]$? It is simply the set $\{a_{0} + a_{1}X + a_{2}X^{2} + \dots + a_{n}X^{n} \mid a_{i} \in \mathbb{F}_{p}, n \in \mathbb{N} \}$. In other words, $\mathbb{F}_{p}[X]$ is the set of polynomials whose coefficients come from $\mathbb{F}_{p}$.

So, for example, we know $\mathbb{F}_{7} = \mathbb{Z}_{7} = \{ [0], [1], [2], [3], [4], [5], [6] \}$. Then some elements in $\mathbb{F}_{7}[X]$ are:

1) $[1] + [3]X^{2} + [5]X^{9}$

2) $[6]X^{2} + [4]X^{3}$

3) $[5]$ (a constant polynomial)

Any polynomial you can think of (remember that polynomials have finite degree, i.e., a finite highest power of $X$) with coefficients in $\mathbb{F}_{7}$ is in $\mathbb{F}_{7}[X]$.

$\mathbb{F}_{7}[X]$ is called the ring of polynomials with coefficients in $\mathbb{F}_{7}$, and this is actually a ring with the addition and multiplication you expect between two polynomials.

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  • $\begingroup$ ohh so F_7=Z_7 because of the fact that 7 is prime... it would be less confusing if they just wrote Z_7 and said it is a field. $\endgroup$ – snowman Nov 17 '14 at 1:09
  • $\begingroup$ @snowman I agree, but now you know! $\mathbb{F}_{7}$ is just $\mathbb{Z}_{7}$. $\endgroup$ – layman Nov 17 '14 at 1:18
  • $\begingroup$ "every element" $\: \mapsto \:$ "every non-zero element" $\;\;\;\;$ $\endgroup$ – user57159 Nov 22 '14 at 22:18
  • $\begingroup$ @RickyDemer What's the significance of this comment? $\endgroup$ – layman Nov 22 '14 at 22:24
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    $\begingroup$ The zero element of $\mathbb{Z}_7$ does not have a multiplicative inverse in $\mathbb{Z}_7$. $\;$ $\endgroup$ – user57159 Nov 22 '14 at 22:25
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$\mathbf{F}_7$ is the finite field with seven elements. Since $7$ is prime, it's isomorphic to the ring $\mathbf{Z}_7$.

If $R$ is a ring, then $R[X]$ is the ring of all polynomials in the indeterminate $X$ whose coefficients are elements in $R$.

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  • $\begingroup$ en.wikipedia.org/wiki/Polynomial_ring $\;$ $\endgroup$ – user57159 Nov 17 '14 at 0:38
  • $\begingroup$ I use the word "the", but we usually don't write $\mathbf{F}_7$ (or even $\mathbf{Z}_7$) to refer to a specific set with $+,-,\cdot$ operations defined on it: usually we don't care which among the many isomorphic rings we are using. And sometimes we might even change our minds and switch back and forth between different representations -- e.g. "integers modulo 7" literally means a particular ring whose elements are equivalence classes of integers, and not a ring whose set of elements are $\{ 0, 1, 2, 3, 4, 5, 6\}$, but everybody what you mean when you write what you did. $\endgroup$ – user14972 Nov 17 '14 at 0:43

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