Question on notation of $\mathbb{F}[t]$

Question

I'm reading an article by Dan Carmon on square-free values of polynomials in function field and I came upon a theorem which I don't fully understand.

It said the following:

Let $\mathcal{P}$ be the set of primes in A (i.e. monic, irreducible polynomials). For any $D\in A$, let $\rho(D):= \#\{a \pmod D:f(a) \equiv 0 \pmod D\}, ||D||:=\#\{a \pmod D\} = q^{\deg D}$, and $c_f:= \prod_{P\in \mathcal{P}} \left( 1 - \frac{\rho(P^2)}{||P^2||}\right)$. Then $$\#\{a \in \mathbb{F}_q[t]:\deg_t(a) < m,\quad f(a)\; \text{square-free}\} = c_fq^m+o(q^m)$$ as $m$ tends to $\infty$.

Before this theorem, the article defined:

• Fix a prime power $q$, let $\mathbb{F}_q$ be the finite field with $q$ elements and let $A = \mathbb{F}_q[t]$ be the ring of polynomials over $\mathbb{F}_q$. $c_f$ is the density function of square free values.

My questions are the following:

• It defined $\mathbb{F}_q$ to be the finite field with $q$ elements, is this equal to $\frac{\mathbb{Z}}{q\mathbb{Z}}$?
• Does $A$ represent the set of all polynomials with coefficients in $\mathbb{F}_q = \{0,1,\ldots,q-1\}$
• Why is the meaning of the set of primes in $A$ equal to monic, irreducible polynomials?
• Since $D \in A$, what does it mean when he talks about $a\pmod D$

Any answers would be very helpful

Answer 1

Edit : the questions are not incorrect but there seems to be a gap between what you want to read and your acquaintance with the field. Maybe you might want to think of $q$ to be prime instead of a prime power, when you read the article.

• No $\mathbb{F}_q$ is not defined as $\frac{\mathbb{Z}}{q\mathbb{Z}}$ (if $q$ is not prime but a prime power, the latter is not even an integral domain whereas the former is a field).

To construct $\mathbb{F}_{p^k}$, you take an irreducible polynomial $P(X)$ of degree $k$ with coefficients in $\frac{\mathbb{Z}}{p\mathbb{Z}}$ and then define : $\mathbb{F}_{p^k}:=\frac{\frac{\mathbb{Z}}{p\mathbb{Z}}[X]}{(P(X))}$. You can also define it as the field generated by the roots of $X^{p^k}-1$ inside the algebraic closure of $\frac{\mathbb{Z}}{p\mathbb{Z}}$.

• $A$ is the set of polynomials with coefficients in the field $\mathbb{F}_{p^k}$.

• $A$ is a commutative unitary ring (you have sum and product for polynomials). It is also an integral domain. So you have a notion of prime elements in $A$ (for the record $p\in A$ is prime if any time $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$). He is simply saying that a set of representative (i.e. modulo. multiplication by an invertible element) prime elements can be chosen to be the set of polynomials that are monic and irreducible.

• $a$ mod $D$ means the natural projection of $a\in A$ to the quotient ring $A/(D)$. If you know the Euclidean division for polynomials, it is kind of harmless if you want to have some intuition to think of "$a$ mod $D$" as the remainder in the Euclidean division of $a$ by $D$. In the same spirit, saying $a=b$ mod $D$ is equivalent to $D$ divides $a-b$.