Counting Number of Irreducibles

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In summary, the conversation discusses the determination of the number of irreducible polynomials over Zp of the form x2 + ax + b and the number of irreducible quadratic polynomials over Zp. The conversation includes a discussion on units in Zp[x] and their relation to irreducibility. The final conclusion is that there are no units in Zp[x] other than 1 and -1, and the number of irreducible polynomials can be determined by subtracting the number of reducible polynomials from the total number of polynomials in Zp[x].
  • #1
e(ho0n3
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Homework Statement
Let p be a prime.
(a) Determine the number of irreducible polynomials over Zp of the form x2 + ax + b.
(b) Determine the number of irreducible quadratic polynomials over Zp.

The attempt at a solution
A nonzero, nonunit polynomial f(x) in Zp[x] is irreducible if it equals the product of two polynomials in Zp[x], one of them being a unit of Zp[x]. But does Zp[x] have any units? I find it hard to imagine that there are polynomials g(x) and h(x) in Zp[x] with g(x)h(x) = 1, unless of course g(x) = h(x) = ±1.
 
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  • #2
e(ho0n3 said:
I find it hard to imagine that there are polynomials g(x) and h(x) in Zp[x] with g(x)h(x) = 1.

just considering the leading terms of g and h will show you this is a trivial statement.

You could also try counting the reducible quadratics, since that is also easy.
 
  • #3
I think I understand: Let axm and bxn be the leading terms of g(x) and h(x). If g(x)h(x) = 1, then abxm+n must equal 0 so ab must equal p but this is impossible. It follows that Zp[x] has no units so the answer both (a) and (b) is 0?
 
  • #4
Oh, and how does counting the reducible quadratics help in counting the irreducible ones?
 
  • #5
How long did you think about that (the counting irreducibles)?
 
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  • #6
Not long. However, something just occurred to me: There are p3 polynomials in Zp[x] so the number of irreducibles is just p3 minus the number of reducibles. Is that right?
 
  • #7
There are *not* p^3 polys in Z_p[x] - there may be p^3 in some subset of that. But it is certainly true that a poly is either reducible or irreducible, and not both, so if you know how many in total there are, and how many of one type, then you know how many of the other there are.

And there are not p^3 quadratics, if that was what you meant, since the leading coefficient has to be non-zero.
 
  • #8
Yes, I did mean p3 quadratics. Sorry about that. So would (p - 1)p2 be a better estimate?

I'm confused though: If Zp[x] doesn't have any units (cf. my second post), how can there be irreducible polynomials?
 
  • #9
In what way does x not satisfy the definition of irreducible? How are you going to write that as a product of two other polynomials of strictly smaller degree. And why are there no units? You even wrote out 2 units in your first post.
 
  • #10
x is irreducible since the only way to write it as a product g(x)h(x) is by setting g(x) = 1 and h(x) = x or vice-versa and 1 is a unit. I understand that.

When I wrote "Zp[x] doesn't have any units", I meant nonconstant units. I can't think of an example unit in Zp[x] that is nonconstant. This is probably why I wrote that there aren't any.
 

1. What are irreducible numbers and why are they important in mathematics?

Irreducible numbers are positive integers that cannot be divided evenly by any other positive integer besides 1 and itself. They are important in mathematics because they are the building blocks of prime numbers, which are essential in many areas of mathematics including number theory and cryptography.

2. How do you determine the number of irreducibles between two given numbers?

To determine the number of irreducibles between two given numbers, you can use the Euler's totient function. This function calculates the number of positive integers less than or equal to a given number that are relatively prime to that number. The result of this function is equal to the number of irreducibles between 1 and the given number.

3. Can all numbers be expressed as a product of irreducibles?

No, not all numbers can be expressed as a product of irreducibles. For example, the number 6 can be expressed as 2 x 3, both of which are irreducible numbers. However, the number 10 cannot be expressed as a product of irreducibles, as it is a composite number with prime factors of 2 and 5.

4. Are there an infinite number of irreducibles?

Yes, there are an infinite number of irreducibles. This is because there are an infinite number of prime numbers, and every prime number is an irreducible number.

5. How are irreducibles used in prime factorization?

Irreducibles are the key components in prime factorization. Every composite number can be expressed as a unique product of irreducibles, known as its prime factorization. This process is important in simplifying fractions, finding common factors, and solving equations involving prime numbers.

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