Finding generators for an ideal of $\Bbb{Z}[x]$

Question

We know that $\Bbb{Z}$ is Noetherian. Hence, we can conclude that $\Bbb{Z}[x]$ is Noetherian, too. Consider the ideal generated by $\langle 2x^2+2,3x^3+3,5x^5+5,…,px^p+p,…\rangle$ for all prime natural numbers $p$. How can I determine a finite number of elements which generate this ideal? Clearly the Euclidean algorithm is not valid in $\Bbb{Z}[x]$.

Thank you

Answer 1

Hint Use the Euclidian Algorithm in $\mathbb{Q}[X]$, then multiply whatever what you get by the common denominator. If needed.

Edit By the extended EA we have

$$3x^3+3=\frac{3}{2}x (2x^2+2)+3-3x$$ $$2x^2+2=-\frac{2}{3}(-3x+3)+4$$

Therefore $$4=2x^2+2+\frac{2}{3}(3x^3+3-\frac{3}{2}x (2x^2+2))\\ =(2x^2+2)(x-1)+\frac{2}{3}(3x^3+3)$$

This shows that $12 \in I$.

Now, for each $p \neq 2,3$ we can write $1=12m+pn$, and hence

$$x^p+1=12(x^p+1)m+n(px^p+1) \in I$$

Using again the Euclidian algorithm in $Q[x]$ we get $$x^7+1=x^2(x^5+1)+(-x^2+1)$$ $$x^5+1=(-x^2+1)(-x^3-x)+x+1$$ $$-x^2+1=(x+1)(-x+1)$$

Thus, $x+1$ is a linear combination of $x^5+1$ and $x^7+1$ with all the coefficients in $Z[X]$. Moreover, as it divides $px^p+p$ for all $p \neq 2$, it is trivial to show

$$I=(2x^2+2, x+1) \,.$$

Now, use again the Euclidean algorithm, for those two polynomials to get $$4=2x^2+2-2(x-1)(x+1)$$

Answer 2

$$ (1 - 11 x - x^2 + 5 x^3 - 5 x^4) (2 x^2 + 2) + 4 (1 + x) (3 x^3 + 3) - 2 (1 - x) (5 x^5 + 5) = 4$$ $$-112 (1 - x^3 - x^5) (5 x^5 + 5) - 80 (x + x^3) (7 x^7 + 7) + \\ 187 (1 - x^3 + x^6 - x^9) (3 x^3 + 3) + 51 x (11 x^{11} + 11) = x+1$$

Conversely, $2x^2+2 = 2(x+1)(x-1)+4$, $px^p+p = p(x+1)(1-x+\dots + x^{p-1})$ for $p$ odd.

Answer 3

This is probably overkill, but it may help you to look into Gröbner bases. They are "good" generators for polynomial ideals, in the sense that they allow you to easily decide ideal membership with a division algorithm, and are minimal in this respect. They are typically defined for polynomial rings over a field, but as I recall, can be calculated for polynomial rings over the integers.

Answer 4

The arguments of the following proof relies on computations within $\mathbb Z[x]$ without any reference to bigger rings.

Let $I=\langle 2x^2+2,3x^3+3,5x^5+5,\dots,px^p+p,\dots\rangle$. Since $$px^p+p=p(x^p+1)=p(x+1)(x^{p-1}-\cdots+1)$$ for any prime $p\ge 3$, we have $$I\subseteq \langle 2x^2+2,x+1\rangle.$$ For $2x^2+2=2(x+1)(x-1)+4$ we get $\langle 2x^2+2,x+1\rangle=\langle 4,x+1\rangle$. Now let's prove that $4,x+1\in I$ and then conclude

$$I=\langle 4,x+1\rangle.$$

Note that $$3(x^3+1)=3(x+1)(x^2-x+1)$$ $$5(x^5+1)=5(x+1)(x^4-x^3+x^2-x+1)=5(x+1)[x^2(x^2-x+1)-x+1],$$ and therefore $15x^2(x^3+1)-15(x^5+1)=15(x^2-1)$. Since $15\cdot2(x^2+1)-30(x^2-1)=60$ we get $60=4\cdot3\cdot 5\in I$. Repeating the same procedure with $11$ and $13$ instead of $3$ and $5$ we get $4\cdot11\cdot13\in I$, and since $(3\cdot5,11\cdot13)=1$ we find $4\in I$.

It remains to prove $x+1\in I$: since $2x^2+2\in I$ and $4\in I$ we get $2x^2-2\in I$ and summing up $4x^2\in I$. Since $3x^3+3\in I$ and $4x^3\in I$ we get (by subtraction) $x^3-3\in I$, so $x^3+1\in I$. From $5x^5+5=5x^2(x^3+1)-5x^2+5\in I$ we get $5x^2-1=(5x^2-5)+4\in I$. Now recall that $4x^2\in I$, so $x^2-1\in I$ and using $x^3+1\in I$ we get $x+1\in I$.