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Difference between revisions of "Cohn's criterion"
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The following proof is due to M. Ram Murty.
 
The following proof is due to M. Ram Murty.
  
−
We start off with a lemma. Let <math>g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. Suppose <math>a_n\geq 1</math>, <math>a-{n-1}\geq 0</math>, and <math>|a_i|\leq H</math>. Then, any complex root of <math>f(x)</math>, <math>\phi</math>, has a non positive real part or satisfies <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>.
+
We start off with a lemma. Let <math>g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]</math>. Suppose <math>a_n\geq 1</math>, <math>|a_i|\leq H</math>.  
 +
Then, any complex root of <math>g(x)</math>, <math>\phi</math>, has a non positive real part or satisfies <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>.
  
 
Proof: If <math>|z|>1</math> and Re <math>z>0</math>, note that:
 
Proof: If <math>|z|>1</math> and Re <math>z>0</math>, note that:
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This means <math>g(z)>0</math> if <math>|z|\geq \frac{1+\sqrt{1+4H}}{2}</math>, so <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>.
 
This means <math>g(z)>0</math> if <math>|z|\geq \frac{1+\sqrt{1+4H}}{2}</math>, so <math>|\phi|<\frac{1+\sqrt{1+4H}}{2}</math>.
  
−
If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible.
+
If <math>b\geq 3</math>, this implies <math>|b-\phi|>1</math> if <math>b\geq 3</math> and <math>f(\phi)=0</math>. Let <math>f(x)=g(x)h(x)</math>. Since <math>f(b)=p</math>, one of <math>|g(b)|</math> and <math>h(b)</math> is 1. WLOG, assume <math>g(b)=1</math>. Let <math>\phi_1, \phi_2,\cdots,\phi_r</math> be the roots of <math>g(x)</math>. This means that <math>|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1</math>. Therefore, <math>f(x)</math> is irreducible.
  
 
If <math>b=2</math>, we will need to prove another lemma:
 
If <math>b=2</math>, we will need to prove another lemma:
  
−
All of the zeroes of <math>f(x)</math> satisfy Re <math>z>\frac{3}{2}</math>.
+
All of the zeroes of <math>f(x)</math> satisfy Re <math>z<\frac{3}{2}</math>.
  
 
Proof: If <math>n=1</math>, then the two polynomials are <math>x</math> and <math>x\pm 1</math>, both of which satisfy our constraint. For <math>n=2</math>, we get the polynomials <math>x^2</math>, <math>x^2\pm x</math>, <math>x^2\pm 1</math>, and <math>x^2\pm x\pm 1</math>, all of which satisfy the constraint. If <math>n\geq 3</math>,  
 
Proof: If <math>n=1</math>, then the two polynomials are <math>x</math> and <math>x\pm 1</math>, both of which satisfy our constraint. For <math>n=2</math>, we get the polynomials <math>x^2</math>, <math>x^2\pm x</math>, <math>x^2\pm 1</math>, and <math>x^2\pm x\pm 1</math>, all of which satisfy the constraint. If <math>n\geq 3</math>,  
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For <math>|z|\geq \frac{3}{2}</math>, then <math>|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1</math>. Therefore, <math>z</math> is not a root of <math>f(x)</math>.
 
For <math>|z|\geq \frac{3}{2}</math>, then <math>|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1</math>. Therefore, <math>z</math> is not a root of <math>f(x)</math>.
  
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However, if Re <math>z<0</math>, we have from our first lemma, that <math>|z|<\frac{1+\sqrt{5}}{2}</math>, so Re <math>z<\frac{1+\sqrt{5}}{2\sqrt{2}}<\frac{3}{2}</math>. Thus we have proved the lemma.
 
  
−
To finish the proof, let <math>f(x)=g(x)h(x)</math>. Since <math>f(2)=p</math>, one of <math>g(2)</math> and <math>h(2)</math> is 1. WLOG, assume <math>g(2)=1</math>. By our lemma, <math>|z-2|>|z-1|</math>. Thus, if <math>\phi_1, \phi_2,\cdots,\phi_r</math> are the roots of <math>g(x)</math>, then <math>|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|>|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)|\leq 1</math>. This is a contradiction, so <math>f(x)</math> is irreducible.
+
To finish the proof, let <math>f(x)=g(x)h(x)</math>. Since <math>f(2)=p</math>, one of <math>g(2)</math> and <math>h(2)</math> is 1. WLOG, assume <math>g(2)=1</math>. By our lemma, <math>|z-2|>|z-1|</math>. Thus, if <math>\phi_1, \phi_2,\cdots,\phi_r</math> are the roots of <math>g(x)</math>, then <math>|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|>|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)| < 1</math>. This is a contradiction, so <math>f(x)</math> is irreducible.
  
  
 
[[Category:Algebra]]
 
[[Category:Algebra]]

Latest revision as of 07:37, 4 March 2021

Let $p$ be a prime number, and $b\geq 2$ an integer. If $\overline{p_np_{n-1}\cdots p_1p_0}$ is the base-$b$ representation of $p$, and $0\leq p_i<b$, then \[f(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots+p_1x+p_0\] is irreducible.

Proof

The following proof is due to M. Ram Murty.

We start off with a lemma. Let $g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]$. Suppose $a_n\geq 1$, $|a_i|\leq H$. Then, any complex root of $g(x)$, $\phi$, has a non positive real part or satisfies $|\phi|<\frac{1+\sqrt{1+4H}}{2}$.

Proof: If $|z|>1$ and Re $z>0$, note that: \[|\frac{g(z)}{z^n}|\geq |a_n+\frac{a_{n-1}}{z}|-H(\frac{1}{|z|^2}+\cdots+\frac{1}{|z|^n})>Re(a_n+\frac{a_{n-1}}{z})-\frac{H}{|z|^2-|z|}\geq 1-\frac{H}{|z|^2-|z|}=\frac{|z|^2-|z|-H}{|z|^2-|z|}\] This means $g(z)>0$ if $|z|\geq \frac{1+\sqrt{1+4H}}{2}$, so $|\phi|<\frac{1+\sqrt{1+4H}}{2}$.

If $b\geq 3$, this implies $|b-\phi|>1$ if $b\geq 3$ and $f(\phi)=0$. Let $f(x)=g(x)h(x)$. Since $f(b)=p$, one of $|g(b)|$ and $h(b)$ is 1. WLOG, assume $g(b)=1$. Let $\phi_1, \phi_2,\cdots,\phi_r$ be the roots of $g(x)$. This means that $|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|>1$. Therefore, $f(x)$ is irreducible.

If $b=2$, we will need to prove another lemma:

All of the zeroes of $f(x)$ satisfy Re $z<\frac{3}{2}$.

Proof: If $n=1$, then the two polynomials are $x$ and $x\pm 1$, both of which satisfy our constraint. For $n=2$, we get the polynomials $x^2$, $x^2\pm x$, $x^2\pm 1$, and $x^2\pm x\pm 1$, all of which satisfy the constraint. If $n\geq 3$, \[|\frac{f(z)}{z^n}|\geq |1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^n})>|1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-\frac{1}{|z|^2(|z|-1)}\]

If Re $z\geq 0$, we have Re $\frac{1}{z^2}\geq 0$, and then \[|\frac{f(z)}{z^n}|>1-\frac{1}{|z|^2(|z|-1)}\] For $|z|\geq \frac{3}{2}$, then $|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}>1$. Therefore, $z$ is not a root of $f(x)$.


To finish the proof, let $f(x)=g(x)h(x)$. Since $f(2)=p$, one of $g(2)$ and $h(2)$ is 1. WLOG, assume $g(2)=1$. By our lemma, $|z-2|>|z-1|$. Thus, if $\phi_1, \phi_2,\cdots,\phi_r$ are the roots of $g(x)$, then $|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|>|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)| < 1$. This is a contradiction, so $f(x)$ is irreducible.

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