Mixing
Is $K[X]$ not a semilocal ring?
$begingroup$
Let $K$ be a field. We will write $K[X]$ to denote the set of all polynomials in one variable over the field $K$ and $mathrm{Maxspec}(K[X])$, the set of all maximal ideals of $K[X].$ Also, we call a ring semilocal, if $|mathrm{Maxspec}(K[X])|<infty$.
I found in a book, that the ring $K[X]$ is not semilocal. But, why does this happen?
Thoughts:
One can prove that
$$mathrm{Maxspec}(K[X])={langle p(X) rangle triangleleft K[X]: p(X) text{ is irreducible over } K[X] }.$$
And this, drives as directly to count the irreducible polynomials over $K[X]$. But, how can we find this cardinality in general?If $K=mathbb{Z}_p$, does this topic help us?
Thank you in advance.
abstract-algebra commutative-algebra semi-simple-rings
asked Jan 5 at 22:30
ChrisChris
836411
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field. We will write $K[X]$ to denote the set of all polynomials in one variable over the field $K$ and $mathrm{Maxspec}(K[X])$, the set of all maximal ideals of $K[X].$ Also, we call a ring semilocal, if $|mathrm{Maxspec}(K[X])|<infty$.
I found in a book, that the ring $K[X]$ is not semilocal. But, why does this happen?
Thoughts:
One can prove that
$$mathrm{Maxspec}(K[X])={langle p(X) rangle triangleleft K[X]: p(X) text{ is irreducible over } K[X] }.$$
And this, drives as directly to count the irreducible polynomials over $K[X]$. But, how can we find this cardinality in general?If $K=mathbb{Z}_p$, does this topic help us?
Thank you in advance.
abstract-algebra commutative-algebra semi-simple-rings
asked Jan 5 at 22:30
ChrisChris
836411
$endgroup$
1
$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
1
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09
add a comment |
$begingroup$
Let $K$ be a field. We will write $K[X]$ to denote the set of all polynomials in one variable over the field $K$ and $mathrm{Maxspec}(K[X])$, the set of all maximal ideals of $K[X].$ Also, we call a ring semilocal, if $|mathrm{Maxspec}(K[X])|<infty$.
I found in a book, that the ring $K[X]$ is not semilocal. But, why does this happen?
Thoughts:
One can prove that
$$mathrm{Maxspec}(K[X])={langle p(X) rangle triangleleft K[X]: p(X) text{ is irreducible over } K[X] }.$$
And this, drives as directly to count the irreducible polynomials over $K[X]$. But, how can we find this cardinality in general?If $K=mathbb{Z}_p$, does this topic help us?
Thank you in advance.
abstract-algebra commutative-algebra semi-simple-rings
asked Jan 5 at 22:30
ChrisChris
836411
$endgroup$
Let $K$ be a field. We will write $K[X]$ to denote the set of all polynomials in one variable over the field $K$ and $mathrm{Maxspec}(K[X])$, the set of all maximal ideals of $K[X].$ Also, we call a ring semilocal, if $|mathrm{Maxspec}(K[X])|<infty$.
I found in a book, that the ring $K[X]$ is not semilocal. But, why does this happen?
Thoughts:
One can prove that
$$mathrm{Maxspec}(K[X])={langle p(X) rangle triangleleft K[X]: p(X) text{ is irreducible over } K[X] }.$$
And this, drives as directly to count the irreducible polynomials over $K[X]$. But, how can we find this cardinality in general?If $K=mathbb{Z}_p$, does this topic help us?
Thank you in advance.
abstract-algebra commutative-algebra semi-simple-rings
abstract-algebra commutative-algebra semi-simple-rings
asked Jan 5 at 22:30
ChrisChris
836411
asked Jan 5 at 22:30
ChrisChris
836411
asked Jan 5 at 22:30
ChrisChris
836411
asked Jan 5 at 22:30
ChrisChris
836411
asked Jan 5 at 22:30
ChrisChris
836411
836411
1
$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
1
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09
add a comment |
1
$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
1
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09
1
1
$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
1
1
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A simpler proof.
We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.
If $k$ is a field, then $k[x]$ has infinitely many irreducibles.
Proof:
Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,ldots,p_n$.
Consider $P=p_1p_2cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)
But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2cdots hat{p_i}cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction.
$k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $blacksquare$
Note
For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.
If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).
General case
For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.
Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever,
but if it continued forever, we get this strictly increasing chain of ideals:
$$(p_0)subsetneq (p_1)subsetneq (p_2)subsetneq cdots subsetneq(p_n)subsetneq cdots,$$
which is impossible. Thus some $p_n$ is irreducible, and $p_nmid p_0$.
That the chain is strictly increasing follows from the fact that if $p_nin (p_{n+1})$, then $p_nmid p_{n+1}$ and $p_{n+1}mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.
answered Jan 5 at 23:03
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
|
show 1 more comment
$begingroup$
This is a very strange claim, and it is very likely to be false.
It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.
Since all $X+a$, $a in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.
Now assume that $K$ is finite.
I claim that $X^{|K|^n}-X$ is the product $Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.
First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u in L longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{circ n}$ is the identity on $L$.
Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 leq |K|^r -1$ hence $d | n$ thus $P | Pi$.
Conversely, let $P$ be an irreducible factor of $Pi$ with degree $d$, and $P neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.
Since $X^{|K|^n}-X$ and $Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $Pi=X^{|K|^n-1}$.
As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.
answered Jan 5 at 22:51
MindlackMindlack
3,11717
$endgroup$
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
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– Chris
Jan 5 at 23:24
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Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A simpler proof.
We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.
If $k$ is a field, then $k[x]$ has infinitely many irreducibles.
Proof:
Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,ldots,p_n$.
Consider $P=p_1p_2cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)
But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2cdots hat{p_i}cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction.
$k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $blacksquare$
Note
For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.
If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).
General case
For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.
Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever,
but if it continued forever, we get this strictly increasing chain of ideals:
$$(p_0)subsetneq (p_1)subsetneq (p_2)subsetneq cdots subsetneq(p_n)subsetneq cdots,$$
which is impossible. Thus some $p_n$ is irreducible, and $p_nmid p_0$.
That the chain is strictly increasing follows from the fact that if $p_nin (p_{n+1})$, then $p_nmid p_{n+1}$ and $p_{n+1}mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.
answered Jan 5 at 23:03
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
|
show 1 more comment
$begingroup$
A simpler proof.
We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.
If $k$ is a field, then $k[x]$ has infinitely many irreducibles.
Proof:
Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,ldots,p_n$.
Consider $P=p_1p_2cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)
But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2cdots hat{p_i}cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction.
$k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $blacksquare$
Note
For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.
If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).
General case
For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.
Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever,
but if it continued forever, we get this strictly increasing chain of ideals:
$$(p_0)subsetneq (p_1)subsetneq (p_2)subsetneq cdots subsetneq(p_n)subsetneq cdots,$$
which is impossible. Thus some $p_n$ is irreducible, and $p_nmid p_0$.
That the chain is strictly increasing follows from the fact that if $p_nin (p_{n+1})$, then $p_nmid p_{n+1}$ and $p_{n+1}mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.
answered Jan 5 at 23:03
jgonjgon
13.6k22041
$endgroup$
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
|
show 1 more comment
$begingroup$
A simpler proof.
We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.
If $k$ is a field, then $k[x]$ has infinitely many irreducibles.
Proof:
Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,ldots,p_n$.
Consider $P=p_1p_2cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)
But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2cdots hat{p_i}cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction.
$k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $blacksquare$
Note
For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.
If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).
General case
For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.
Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever,
but if it continued forever, we get this strictly increasing chain of ideals:
$$(p_0)subsetneq (p_1)subsetneq (p_2)subsetneq cdots subsetneq(p_n)subsetneq cdots,$$
which is impossible. Thus some $p_n$ is irreducible, and $p_nmid p_0$.
That the chain is strictly increasing follows from the fact that if $p_nin (p_{n+1})$, then $p_nmid p_{n+1}$ and $p_{n+1}mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.
answered Jan 5 at 23:03
jgonjgon
13.6k22041
$endgroup$
A simpler proof.
We can prove that there are infinitely many irreducibles by adapting Euclid's proof from the integers.
If $k$ is a field, then $k[x]$ has infinitely many irreducibles.
Proof:
Suppose there are not infinitely many irreducible polynomials, suppose there are only $p_1,ldots,p_n$.
Consider $P=p_1p_2cdots p_n+1$. Some irreducible $p_i$ must divide it since it has positive degree and is therefore not a unit. (See the note)
But then we have $P=p_iq+1=p_ik$ where $q=p_1p_2cdots hat{p_i}cdots p_n$ and $k$ is the polynomial such that $P=p_ik$, so $p_i(k-q)=1$, and $p_i$ is a unit. Contradiction.
$k[x]$ has infinitely many irreducibles. (This proof of infinitely many irreducibles generalizes to $k$ any Noetherian domain) $blacksquare$
Note
For the fewest assumptions, we can prove that any positive degree polynomial over a field is divisible by an irreducible by induction.
If $P$ has positive degree and is not irreducible itself, then $P=ab$ for $ab$ both nonunits, and hence both of positive degree (necessarily less than that of $P$). Thus one of $a$ or $b$ is divisible by an irreducible by induction. (Note that the base case here is if $P$ is irreducible).
General case
For the case of $k$ a Noetherian domain, we use the fact that $k$ Noetherian implies $k[x]$ is Noetherian, and the fact that in any Noetherian domain any nonunit is divisible by some irreducible, which is proved in the following manner.
Let $p_0$ be a nonunit. If $p_0$ is irreducible, then we are done. Otherwise $p_0=p_1a_1$ with $p_1,a_1$ nonunits. Now if $p_1$ is irreducible, we are done. Otherwise, we repeat to get $p_1=p_2a_2$, and $p_n=p_{n+1}a_{n+1}$. Then either some $p_{n+1}$ is irreducible, or this continues forever,
but if it continued forever, we get this strictly increasing chain of ideals:
$$(p_0)subsetneq (p_1)subsetneq (p_2)subsetneq cdots subsetneq(p_n)subsetneq cdots,$$
which is impossible. Thus some $p_n$ is irreducible, and $p_nmid p_0$.
That the chain is strictly increasing follows from the fact that if $p_nin (p_{n+1})$, then $p_nmid p_{n+1}$ and $p_{n+1}mid p_n$, which in a domain would imply that $a_{n+1}$ is a unit, which would be a contradiction.
answered Jan 5 at 23:03
jgonjgon
13.6k22041
edited Jan 5 at 23:14
answered Jan 5 at 23:03
jgonjgon
13.6k22041
answered Jan 5 at 23:03
jgonjgon
13.6k22041
answered Jan 5 at 23:03
jgonjgon
13.6k22041
13.6k22041
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
|
show 1 more comment
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Yes, this is indeed far simpler than my argument. I had never thought of this...
$endgroup$
– Mindlack
Jan 5 at 23:10
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
Thank you for your answer. Simple and elegant.
$endgroup$
– Chris
Jan 5 at 23:15
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@Mindlack The proof you gave is also very important and useful though. Iirc, a counting argument based on it proves that finite fields have irreducible polynomials of every degree, which is a stronger result than what I proved.
$endgroup$
– jgon
Jan 5 at 23:17
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
$begingroup$
@jgon Can we claim, istead of your note, that $K[X]$ is a UFD, so $p(X)=cp_1(X)^{n_1} cdots p_k(X)^{n_k}$, for some $kin mathbb{N}$ and $n_1,...,n_kin mathbb{Z}^+, cin K^*$?
$endgroup$
– Chris
Jan 5 at 23:43
1
1
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
$begingroup$
@Chris, sure, there are many proofs of that fact (which is why I put it in the note at the end, since you can substitute any of many proofs for that fact)
$endgroup$
– jgon
Jan 6 at 0:02
|
show 1 more comment
$begingroup$
This is a very strange claim, and it is very likely to be false.
It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.
Since all $X+a$, $a in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.
Now assume that $K$ is finite.
I claim that $X^{|K|^n}-X$ is the product $Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.
First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u in L longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{circ n}$ is the identity on $L$.
Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 leq |K|^r -1$ hence $d | n$ thus $P | Pi$.
Conversely, let $P$ be an irreducible factor of $Pi$ with degree $d$, and $P neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.
Since $X^{|K|^n}-X$ and $Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $Pi=X^{|K|^n-1}$.
As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.
answered Jan 5 at 22:51
MindlackMindlack
3,11717
$endgroup$
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
add a comment |
$begingroup$
This is a very strange claim, and it is very likely to be false.
It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.
Since all $X+a$, $a in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.
Now assume that $K$ is finite.
I claim that $X^{|K|^n}-X$ is the product $Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.
First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u in L longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{circ n}$ is the identity on $L$.
Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 leq |K|^r -1$ hence $d | n$ thus $P | Pi$.
Conversely, let $P$ be an irreducible factor of $Pi$ with degree $d$, and $P neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.
Since $X^{|K|^n}-X$ and $Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $Pi=X^{|K|^n-1}$.
As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.
answered Jan 5 at 22:51
MindlackMindlack
3,11717
$endgroup$
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
add a comment |
$begingroup$
This is a very strange claim, and it is very likely to be false.
It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.
Since all $X+a$, $a in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.
Now assume that $K$ is finite.
I claim that $X^{|K|^n}-X$ is the product $Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.
First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u in L longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{circ n}$ is the identity on $L$.
Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 leq |K|^r -1$ hence $d | n$ thus $P | Pi$.
Conversely, let $P$ be an irreducible factor of $Pi$ with degree $d$, and $P neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.
Since $X^{|K|^n}-X$ and $Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $Pi=X^{|K|^n-1}$.
As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.
answered Jan 5 at 22:51
MindlackMindlack
3,11717
$endgroup$
This is a very strange claim, and it is very likely to be false.
It states basically that the ring $K[X]$ is semilocal iff it has finitely many irreducible monic polynomials.
Since all $X+a$, $a in K$, are monic irreducible polynomials, if $K[X]$ is semilocal, then $K$ is finite.
Now assume that $K$ is finite.
I claim that $X^{|K|^n}-X$ is the product $Pi$ of all monic irreducible polynomials over $K$ such that their degree divides $n$.
First, let $P$ denote some irreducible factor of $X^{|K|^n}-X$, let $d$ be its degree. Then $L=K[X]/(P)$ is a field extension of degree $d$ of $K$. Let $F : u in L longmapsto u^{|K|}$, it is a field automorphism of $L$, and $F^{circ n}$ is identity on $K$ and the image of $X$ in $L$ (Because $P|X^{|K|^n}-X$). These elements span $L$ as a ring, thus $F^{circ n}$ is the identity on $L$.
Now, let $r$ be the gcd of $n$ and $d$, then $|K|^r-1$ is the gcd of $|K|^n-1$ and $|K|^d-1=|L^*|$. Since every element of $L^*$ is a root of $X^{|K|^n-1}-1$ and of $X^{|K|^d-1}-1$, the polynomial $X^{|K|^r-1}-1$ vanishes at every element of $L^*$ so has at least $|K|^d-1$ roots. Thus $|K|^d-1 leq |K|^r -1$ hence $d | n$ thus $P | Pi$.
Conversely, let $P$ be an irreducible factor of $Pi$ with degree $d$, and $P neq X$. Then $L=K[X]/(P)$ is a field extension of $K$ with degree $d$, so $L^*$ is a group of cardinality $|K|^d-1$, thus the identity $X^{|K|^d-1}=1$ holds in $L$, thus $X^{|K|^n-1}=1$ in $L$ as well, which translates to $P | X^{|K|^n-1}-1$ in $K[X]$.
Since $X^{|K|^n}-X$ and $Pi$ have the same irreducible factors, are monic, and do not have double factors (check the derivative for the first one), $Pi=X^{|K|^n-1}$.
As a consequence: there exists a monic irreducible polynomial in $K$ of degree $2^n$ for every $n$. Thus there are infinitely many irreducible polynomials over $K$ and $K[X]$ is not semilocal.
answered Jan 5 at 22:51
MindlackMindlack
3,11717
answered Jan 5 at 22:51
MindlackMindlack
3,11717
answered Jan 5 at 22:51
MindlackMindlack
3,11717
answered Jan 5 at 22:51
MindlackMindlack
3,11717
3,11717
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
add a comment |
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Thank you for your nice answer. Let me please study it carefully to understand it.
$endgroup$
– Chris
Jan 5 at 23:24
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
$begingroup$
Could you please explain this point: What do you mean $F : u in L longmapsto u^{|K|}$ and what is exactly $F^{circ n}$?
$endgroup$
– Chris
Jan 6 at 0:15
1
1
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
$begingroup$
I consider the application $F$ from $L$ to $L$ defined by the formula. It is a field automorphism. And $F^{circ n}=overbrace{F circ F ldots circ F}^{n text{ times}}$.
$endgroup$
– Mindlack
Jan 6 at 0:22
add a comment |
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$begingroup$
If $K$ is infinite, then $)X-a)$ give us infinitely many prime ideals. If $K$ is finite....
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:39
$begingroup$
Thank you for your answer. Ok, if $K$ is infinite, then $X-a in K[X]$ is irreducible, for all $ain K$. So, there are infinitely many $langle X-a rangle $, and thus $K[X]$ is not semilocal. But, what if $|K|<infty$?
$endgroup$
– Chris
Jan 5 at 22:45
$begingroup$
If $K$ is finite, one has to work a little harder to get infinitely many prime ideals.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:46
1
$begingroup$
That there are infinetly many irreducible polynomials is basically equivalent to the classification of finite fields.
$endgroup$
– RghtHndSd
Jan 5 at 22:48
$begingroup$
In fact $R[x]$ is not semi-local for any ring with identity, even if $R$ has zero divisors. Let $M$ be a maximal ideal of $R$. Then $R[x]/M[x] cong (R/M)[x]$ is a polynomial ring over a field. As @jgon shows, $(R/M)[x]$ will thus have infinitely many maximal ideals, and the ideal correspondence for quotient rings gives us infinitely many maximal ideals of $R[x]$ containing $M[x]$.
$endgroup$
– Badam Baplan
Jan 6 at 15:09