Let k be a finite field. Is it true that the number of irreducible polynomials in k[x] is also finite?
$begingroup$
I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?
field-theory finite-fields irreducible-polynomials
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
$endgroup$
add a comment |
$begingroup$
I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?
field-theory finite-fields irreducible-polynomials
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
$endgroup$
$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07
add a comment |
$begingroup$
I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?
field-theory finite-fields irreducible-polynomials
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
$endgroup$
I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?
field-theory finite-fields irreducible-polynomials
field-theory finite-fields irreducible-polynomials
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
asked Dec 10 '18 at 21:52
Jon D.Jon D.
164
164
$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07
add a comment |
$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07
$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)
In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
$endgroup$
add a comment |
$begingroup$
Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.
Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.
Therefore $K$ is algebraically closed.
Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
$$
(x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
$$
has no root in $K$.
Contradiction.
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.
For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
$endgroup$
add a comment |
$begingroup$
No, there exist irreducible polynomial in every degree, due to the following formula:
Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
$$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
$endgroup$
add a comment |
$begingroup$
So after more reading, I believe I can show it is false by a counter-example.
Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.
Therefore, the initial statement is false.
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
$endgroup$
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)
In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
$endgroup$
add a comment |
$begingroup$
One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)
In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
$endgroup$
add a comment |
$begingroup$
One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)
In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
$endgroup$
One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)
In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
answered Dec 10 '18 at 22:05
SladeSlade
25.1k12665
25.1k12665
add a comment |
add a comment |
$begingroup$
Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.
Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.
Therefore $K$ is algebraically closed.
Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
$$
(x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
$$
has no root in $K$.
Contradiction.
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.
Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.
Therefore $K$ is algebraically closed.
Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
$$
(x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
$$
has no root in $K$.
Contradiction.
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.
Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.
Therefore $K$ is algebraically closed.
Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
$$
(x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
$$
has no root in $K$.
Contradiction.
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
$endgroup$
Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.
Suppose $K'$ is an algebraic extension field of $K$; if $bin K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $bin K$.
Therefore $K$ is algebraically closed.
Let $K={a_1=0,a_2=1,a_3,dots,a_n}$. The polynomial
$$
(x-a_1)(x-a_2)(x-a_3)dots(x-a_n)+1
$$
has no root in $K$.
Contradiction.
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
answered Dec 10 '18 at 22:38
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.
For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
$endgroup$
add a comment |
$begingroup$
This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.
For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
$endgroup$
add a comment |
$begingroup$
This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.
For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
$endgroup$
This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.
For every finite field extension $Ksupseteq F$, there is an irreducible $pin F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
answered Dec 10 '18 at 22:21
tomasztomasz
23.6k23479
23.6k23479
add a comment |
add a comment |
$begingroup$
No, there exist irreducible polynomial in every degree, due to the following formula:
Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
$$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
$endgroup$
add a comment |
$begingroup$
No, there exist irreducible polynomial in every degree, due to the following formula:
Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
$$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
$endgroup$
add a comment |
$begingroup$
No, there exist irreducible polynomial in every degree, due to the following formula:
Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
$$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
$endgroup$
No, there exist irreducible polynomial in every degree, due to the following formula:
Let $p$ be a prime number, $q=p^n$. In $ mathbf F_q[X]$, we have the factorisation:
$$ X^{q^n}-X=prod_{dmid n}prod_{:deg P=d,\text{ irreducible}}mkern-12mu P$$
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
answered Dec 10 '18 at 22:27
BernardBernard
120k740113
120k740113
add a comment |
add a comment |
$begingroup$
So after more reading, I believe I can show it is false by a counter-example.
Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.
Therefore, the initial statement is false.
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
$endgroup$
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
add a comment |
$begingroup$
So after more reading, I believe I can show it is false by a counter-example.
Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.
Therefore, the initial statement is false.
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
$endgroup$
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
add a comment |
$begingroup$
So after more reading, I believe I can show it is false by a counter-example.
Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.
Therefore, the initial statement is false.
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
$endgroup$
So after more reading, I believe I can show it is false by a counter-example.
Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ.
Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.
Therefore, the initial statement is false.
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
answered Dec 10 '18 at 23:51
Jon D.Jon D.
164
164
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
add a comment |
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
$begingroup$
That approach is already in this older dupe.
$endgroup$
– Bill Dubuque
Dec 11 '18 at 0:30
add a comment |
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$begingroup$
Here: math.stackexchange.com/q/80389/2838
$endgroup$
– Joe Johnson 126
Dec 10 '18 at 22:03
$begingroup$
For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $,(x^n-1)/(x-1) $
$endgroup$
– Bill Dubuque
Dec 10 '18 at 22:07