A real polynomial of degree more than or equal to 3 is reducible, but does it necessarily have a real zero?
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Takumi Murayama says "Every polynomial in $mathbb R[x]$ of degree at least 3 has a real root, and therefore is not irreducible". I think I understand why it is not irreducible, but what's the real root of $f(x)=(x^2+1)(x^2+2)$?
If he is right, then why?
If it is wrong, then what is probably meant? I think this has something to do with complex roots in pairs.
abstract-algebra polynomials ideals irreducible-polynomials maximal-and-prime-ideals
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show 3 more comments
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Takumi Murayama says "Every polynomial in $mathbb R[x]$ of degree at least 3 has a real root, and therefore is not irreducible". I think I understand why it is not irreducible, but what's the real root of $f(x)=(x^2+1)(x^2+2)$?
If he is right, then why?
If it is wrong, then what is probably meant? I think this has something to do with complex roots in pairs.
abstract-algebra polynomials ideals irreducible-polynomials maximal-and-prime-ideals
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4
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I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
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– lulu
Dec 22 '18 at 12:06
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Yes, lulu is right; see this post.
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– Dietrich Burde
Dec 22 '18 at 12:08
1
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The link from Murayama you attach is extremely long. On what page does that quote appear?
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– lulu
Dec 22 '18 at 12:09
3
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The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
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– Michael Hoppe
Dec 22 '18 at 12:09
1
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@JackBauer Well, in my opinion a comment is sufficient.
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– Michael Hoppe
Dec 25 '18 at 20:44
|
show 3 more comments
$begingroup$
Takumi Murayama says "Every polynomial in $mathbb R[x]$ of degree at least 3 has a real root, and therefore is not irreducible". I think I understand why it is not irreducible, but what's the real root of $f(x)=(x^2+1)(x^2+2)$?
If he is right, then why?
If it is wrong, then what is probably meant? I think this has something to do with complex roots in pairs.
abstract-algebra polynomials ideals irreducible-polynomials maximal-and-prime-ideals
$endgroup$
Takumi Murayama says "Every polynomial in $mathbb R[x]$ of degree at least 3 has a real root, and therefore is not irreducible". I think I understand why it is not irreducible, but what's the real root of $f(x)=(x^2+1)(x^2+2)$?
If he is right, then why?
If it is wrong, then what is probably meant? I think this has something to do with complex roots in pairs.
abstract-algebra polynomials ideals irreducible-polynomials maximal-and-prime-ideals
abstract-algebra polynomials ideals irreducible-polynomials maximal-and-prime-ideals
asked Dec 22 '18 at 12:02
user198044
4
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I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
$endgroup$
– lulu
Dec 22 '18 at 12:06
$begingroup$
Yes, lulu is right; see this post.
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– Dietrich Burde
Dec 22 '18 at 12:08
1
$begingroup$
The link from Murayama you attach is extremely long. On what page does that quote appear?
$endgroup$
– lulu
Dec 22 '18 at 12:09
3
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The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
$endgroup$
– Michael Hoppe
Dec 22 '18 at 12:09
1
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@JackBauer Well, in my opinion a comment is sufficient.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 20:44
|
show 3 more comments
4
$begingroup$
I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
$endgroup$
– lulu
Dec 22 '18 at 12:06
$begingroup$
Yes, lulu is right; see this post.
$endgroup$
– Dietrich Burde
Dec 22 '18 at 12:08
1
$begingroup$
The link from Murayama you attach is extremely long. On what page does that quote appear?
$endgroup$
– lulu
Dec 22 '18 at 12:09
3
$begingroup$
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
$endgroup$
– Michael Hoppe
Dec 22 '18 at 12:09
1
$begingroup$
@JackBauer Well, in my opinion a comment is sufficient.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 20:44
4
4
$begingroup$
I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
$endgroup$
– lulu
Dec 22 '18 at 12:06
$begingroup$
I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
$endgroup$
– lulu
Dec 22 '18 at 12:06
$begingroup$
Yes, lulu is right; see this post.
$endgroup$
– Dietrich Burde
Dec 22 '18 at 12:08
$begingroup$
Yes, lulu is right; see this post.
$endgroup$
– Dietrich Burde
Dec 22 '18 at 12:08
1
1
$begingroup$
The link from Murayama you attach is extremely long. On what page does that quote appear?
$endgroup$
– lulu
Dec 22 '18 at 12:09
$begingroup$
The link from Murayama you attach is extremely long. On what page does that quote appear?
$endgroup$
– lulu
Dec 22 '18 at 12:09
3
3
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The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
$endgroup$
– Michael Hoppe
Dec 22 '18 at 12:09
$begingroup$
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
$endgroup$
– Michael Hoppe
Dec 22 '18 at 12:09
1
1
$begingroup$
@JackBauer Well, in my opinion a comment is sufficient.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 20:44
$begingroup$
@JackBauer Well, in my opinion a comment is sufficient.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 20:44
|
show 3 more comments
1 Answer
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From the comments
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09
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From the comments
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09
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add a comment |
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From the comments
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09
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add a comment |
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From the comments
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09
$endgroup$
From the comments
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial." – Michael Hoppe Dec 22 at 12:09
edited Dec 29 '18 at 12:49
answered Dec 29 '18 at 12:37
user198044
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I believe the writer intended to add that the degree was odd. Otherwise, the claim is false for the reason you report. There are polynomials of arbitrarily high even degree with no root, $p_n(x)=(x^2+1)^n$ has degree $2n$ and has no real root, for example.
$endgroup$
– lulu
Dec 22 '18 at 12:06
$begingroup$
Yes, lulu is right; see this post.
$endgroup$
– Dietrich Burde
Dec 22 '18 at 12:08
1
$begingroup$
The link from Murayama you attach is extremely long. On what page does that quote appear?
$endgroup$
– lulu
Dec 22 '18 at 12:09
3
$begingroup$
The author's claim (on p. 39) is wrong. What he meant is that every polynomial of degree three or more has irreducible factors which are at most quadratic: "So, every maximal ideal is of the form (f) for f a linear polynomial or an irreducible quadratic polynomial."
$endgroup$
– Michael Hoppe
Dec 22 '18 at 12:09
1
$begingroup$
@JackBauer Well, in my opinion a comment is sufficient.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 20:44