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I am wondering if we can find an irreducible polynomial $g(x)$ in $\mathbb{Z}[x]$ such that

  • The constant term, $c(g)=\pm 1$ and the leading coefficient $\ell(g)=\pm 1$,
  • the ideal generated by $g(x)$ and $5x+7$ is the ring $\mathbb{Z}[x]$, that is, $(g(x),5x+7)=1$, in other words: $g(-7/5)=\pm 1$,
  • the ideal generated by $g(x)$ and $2x-3$ is the ring $\mathbb{Z}[x]$, that is, $(g(x), 2x-3)=1$, in other words: $g(3/2)=\pm 1$.


PS: You can change $5x+7$ and $2x-3$ with any polynomials such that their constant terms and the leading coefficients are not units in $\mathbb{Z}$.

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If $g \in \mathbb{Z}[x]$ is irreducible, then $\gcd(f,g) = 1$ for all $f \in \mathbb{Z}[x] - \{g, 0\}$. – user2468 Mar 7 '12 at 1:12
Seems trivial, but how about $x^2 + x + 1$?! – user2468 Mar 7 '12 at 1:52
@J.D., is 1 in the ideal of ${\bf Z}[x]$ generated by $x^2+x+1$ and $2x-3$? We're not in ${\bf Q}[x]$. – Gerry Myerson Mar 7 '12 at 2:13
@GerryMyerson, it certainly isn't. $\mathbb{Z}[x]/(x^2+x+1)$ is isomorphic to $\mathbb{Z}[\omega]$, the ring of integers of the 3rd cyclotomic field, and $2x-3$ corresponds to $2\omega-3$ under this isomorphism. The group of units in there is torsion, and $2\omega-3$ is not one of them, so the ideal generated by $(2\omega-3)$ cannot be all of $\mathbb{Z}[\omega]$. In fact, the norm of $(2\omega-3)$ down to $\mathbb{Q}$ is 16. – Alex B. Mar 7 '12 at 2:22
I would like to thank you for thinking about this question. First of all, I forgot to write that c(f) is the constant term and l(f) is the leading term in the first bullet and this is not a homework question. I have been trying to find these special irreducible polynomials to give an example in my research. I could find infinitely many irreducible polynomials over the integers if I change 5x+7 with x+7 and 2x-3 with x-3. But I don't know if we have such an irreducible polynomial in this case. – eileendavid82 Mar 8 '12 at 1:08

$g(x)=x^2-2$ works. (I'm taking $c$ to be the content, and $\ell$ to be the leading coefficient). Verification let to the reader, in case this is homework.

EDITed in the light of comments and edits to the question statement:

If non-constant $g$ has integer coefficients and leading coefficient $\pm1$, then $g(3/2)=\pm1$ is obviously impossible. However, $g(3/2)=\pm1$ has nothing to do with the ideal generated by $g$ and $2x-3$.

For example, if $g(x)=x^2-x-1$, then $g$ satisfies all your conditions: it is irreducible, has integer coefficients, has leading coefficient and constant term $\pm1$, and with $2x-3$ generates the ring ${\bf Z}[x]$, as is evident from $$(-4)(x^2-x-1)+(2x+1)(2x-3)=1$$

I don't know whether there is a polynomial that satisfies your conditions simultaneously for $2x-3$ and $5x+7$, but if I find one, I'll let you know.

Further EDIT: You say I can change the polynomials. If I change the $5x+7$ to $5x-8$ then $x^2-x-1$ will solve your problem.

Even more EDIT: I'm now confident (but not 100% certain, since I haven't carried out all the calculations) that there is a polynomial $g$ of degree 6 with integer coefficients, leading coefficient 1, constant term $-1$, irreducible over the rationals, such that $1$ is in both the ideals $(g,5x+7)$ and $(g,2x-3)$.

The condition on the ideals will be satisfied if $5^6g(-7/5)=-1$ and $2^6g(3/2)=-1$. Let $$g(x)=x^6+ax^5+bx^4+cx^3+dx^2+ex-1$$ Then we get the two equations in $5$ unknowns, $$7^6-7^55a+7^45^2b-7^35^3c+7^25^4d-(7)5^5e-5^6=-1$$ and $$3^6+3^52a+3^42^2b+3^32^3c+3^22^4d+(3)2^5e-2^6=-1$$ Move the constant terms to the right side of the equations, divide the 1st one by $-35$ (note that $7^6-5^6+1$ is a multiple of $35$) and the second one by $6$ ($3^6-2^6+1$ is a multiple of $6$), and you have two linear equations in $5$ unknowns, with no modular obstacle to a solution.

Now I wave my hands a little and say there must be infinitely many integer solutions to this pair of equations, including infinitely many with $g$ irreducible. In any event, it should not be hard to find one such solution.

Please get back to me if there are any questions about this.

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Sorry but I forgot to mention that c is the constant term and \ell is the leading coefficient. – eileendavid82 Mar 8 '12 at 1:18
@Gerry: Unless I've made a horrible mistake, your system of equations has no solution modulo 29. Using wolframalpha to solve for $d$ and $e$ gives an expression with $29$ in the denominator, and a numerator that is a non-zero constant modulo $29$. – Hurkyl Mar 11 '12 at 2:24
@Hurkyl, not sure I understand. How are you solving for $d$ and $e$, when there's also $a$, $b$, and $c$? – Gerry Myerson Mar 11 '12 at 4:26
@Gerry: Solving for $d$ and $e$ in terms of $a$, $b$, and $c$. – Hurkyl Mar 11 '12 at 11:49
@Hurkyl, so, the numerator depends on $a$, $b$, and $c$, but it's a non-zero constant mod 29? Very curious. – Gerry Myerson Mar 11 '12 at 11:57

This is not an answer yet....

Suppose $g$ has degree $d$, and let $G(x,y) = y^d g(x/y)$ be the homogenization of $g$. Write

$$ G(x,y) = \sum_{k=0}^d c_k x^k y^{d-k} $$

The constraints of your problem are that

  • $G(1, 0) = \pm 1$
  • $G(0, 1) = \pm 1$
  • $G(-7, 5) = \pm 1$
  • $G(3, 2) = \pm 1$

By checking various moduli, we have the equations

  • $G(3, 2) = 2^d G(0,1) \pmod 3$
  • $G(-7, 5) = (-7)^d G(1, 0) \pmod 5$
  • $G(-7, 5) = 5^d G(0, 1) \pmod 7$
  • $G(3, 2) = 12^d G(-7, 5) \pmod {29}$

With our constraint that all of these evaluations must be $\pm 1$, you can show that $d \equiv 0 \pmod {12}$, and that all of the signs must be the same. (I choose positive)

I haven't had any success beyond this point....

EDIT: Ah, I've found a solution:

I was doing something similar to what Gerry was doing last night, but apparently I was doing something wrong (or wolframalpha wasn't doing what I though it was doing), since I sat down and wrote a python program to do the degree 12 case, and got

$$1 - 7172241325351892585589 x - 89487744748539793040 x^{10} + 184036134006187327480 x^{11} + x^{12}$$

as a solution. Here's a check that the resultant is, indeed 1:

and the extended gcd calculation:

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Excellent work. – Gerry Myerson Mar 19 '12 at 10:11

Hint $\rm\ \ (g(x),\:3-2x)\: =\: 1\:$ in $\rm\:\mathbb Z[x]\iff 3-2x\ |\ 1\:$ in $\rm\: R\: =\: \mathbb Z[x]/g(x).\:$

If $\rm\:g(x)\: =\: x^2 - d\:$ then $\rm\:R = \mathbb Z[\sqrt{d}]\:$ so $\rm\:3-2\sqrt{d}\ |\ 1\:\Rightarrow\: 9-4d\: =\: \pm1\:$ by taking norms.

Similarly $\rm\:7+5\sqrt{d}\ |\ 1\:\Rightarrow\: 49-25d\ =\: \pm 1.$

So it suffices to solve $\rm\: 9-4d\: =\: \pm1,\ \ 49-25d\: =\: \pm 1,\:$ yielding $\rm\:d = 2,\ \ g\: =\: x^2 -2.$

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Nice. Not clear where the "If $g(x)=x^2-d$ comes from. Also, the problem can, of course, be done without recourse to arithmetic in quadratic number fields. – Gerry Myerson Mar 7 '12 at 5:33
c(f)=constant term, l(f)=leading coefficient. – eileendavid82 Mar 8 '12 at 1:11

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