Let $S= \mathbb R[x]$ denote the polynomial ring in one variable over the field $\mathbb R$ of real numbers. Find a monic polynomial of least degree in $S$ that is a square root of $−4$ modulo the ideal $I$ generated by $(x^2 + 1)^2$.

Let $p(x)$ be a monic polynomial that is a square root of $−4$ modulo the ideal $I$.

Then ${(p(x))}^2 = -4 + \langle(x^2 + 1)^2\rangle$.

$p(x)$ is of the form $a_0 + a_1x + a_2x^2 + x^3$. Thus, $(a_0 +a_1x + a_2x^2 + x^3)^2 = -4 + \langle (x^2 + 1)^2\rangle$.

Solving this rigorously (reducing and equating) may get such $p(x)$.

But Is there any other simple way?

  • 4
    $\begingroup$ $(2x)^2\equiv-4\pmod{\langle x^2+1\rangle}$, so I suggest that you try an "lift" that to a solution modulo the square. That is, an ansatz of the form $p(x)=2x+a(x)(x^2+1)$ should work. Here $a(x)$ is an at most linear polynomial to be determined. $\endgroup$ Sep 8 '21 at 9:48
  • $\begingroup$ Its is mod$<(x^2+1)^2>$. There is a square term. $\endgroup$
    – Nick Diaz
    Sep 8 '21 at 10:09
  • 1
    $\begingroup$ We have plenty of threads discussing Hensel lifting. $\endgroup$ Sep 8 '21 at 11:23

The basis $1,x,x^2,x^3$ has no relation to $(x^2+1)^2$, which is the polynomial which you are trying to go modulo. Instead, what is usually done is traditionally called "lifting" as suggested in the comments by Jyrki. (So it wasn't misread, it was a hint).

The idea of lifting is simple : if an equation is true modulo $p(x)$, then it's also true modulo $q(x)$ where $q$ is a factor of $p$. Therefore, it should be possible to solve the equation modulo $q(x)$ and then "lift" it to a solution modulo $p(x)$, using the solution obtained modulo $q(x)$.

In this scenario, $x^2+1$ is a factor of $(x^2+1)^2$, so you want to first solve this equation modulo $x^2+1$, and then modulo $(x^2+1)^2$, by lifting.

Let's first solve $q(x)^2 = -4 \pmod{x^2+1}$. Now, here the degree of the polynomial $q$ can be chosen to be at most $1$ i.e. $q(x) = ax+b$ for some $a,b$. Then, note that $(ax+b)^2 = a^2x^2+2abx+b^2$ , which modulo $x^2+1$ becomes $2abx + (b^2-a^2)$. Thus, $2abx + (b^2-a^2) = 4 \pmod{x^2+1}$, hence $2ab=0$ and $b^2-a^2 = 4$. We easily get $b=0,a=\pm 2$ from here. Sticking with $a=2$ gives us one of the solutions, $q(x) = 2x$. (And the set of all solutions would be $2x$ or $-2x$ modulo $x^2+1$).

Now, suppose that $p(x)$ satisfies $p(x)^2 = -4 \pmod{(x^2+1)^2}$. Then, $p$ also satisfies $p(x)^2 = -4 \pmod{x^2+1}$, but then we've already seen that one solution to this is $q(x)$! Therefore, there could be a solution $p(x) \equiv q(x) \equiv 2x\pmod{x^2+1}$. In fact, there is one. Similarly, there will also be a solution for $p(x) \equiv -2x \pmod{x^2+1}$, as I urge you to calculate (later on, you can see the similarity between the solution you get from here, and the solution I'm about to derive : they will be $-1$ times each other).

Thus, suppose that $p(x) = (ax+b)(x^2+1) - 2x$ for some $a,b$ (following the realization that $p$ has degree at most $3$). We need this to satisfy $p(x)^2 = -4 \pmod{(x^2+1)^2}$. While squaring $p(x)$, the entire $(ax+b)(x^2+1)$ term squares to something that is $0$ modulo $(x^2+1)^2$. Neglecting this gives $$ 2(ax+b)(x^2+1)(-2x) + (4x^2) = -4 \pmod{(x^2+1)^2} $$

which ... we don't need to expand and make life complicated! Why? Because taking the $-4$ to the other side : $$ 2(ax+b)(x^2+1)(-2x) + (4x^2 +4) = 0 \pmod{(x^2+1)^2} \\ \implies 2(ax+b)(-2x)(x^2+1)+4(x^2+1)=0 \pmod{(x^2+1)^2} \\ \implies (x^2+1) \left[(2(ax+b)(-2x) + 4)\right] = 0 \pmod{(x^2+1)^2} $$

which happens if and only if $$ (2(ax+b)(-2x) + 4) = 0 \pmod{x^2+1} $$

and expanding this out gives us $-4ax^2-4bx + 4 = 0 \pmod{x^2+1}$, which then simplifies to $-4bx + 4a+4 =0 \pmod{x^2+1}$. As the LHS is linear, this can occur only if $b = 0$ and $a = -1$. Thus, the final $p(x)$ is $-x(x^2+1) - 2x = -x^3-3x$.

To confirm this, one can check that $(-x^3-3x)^2 =-4 \pmod{(x^2+1)^2}$.

The other solution, will in fact be $x^3+3x$, which you can note is a different element modulo $(x^2+1)^2$. Our analysis confirms that these are the only two solutions.

  • 2
    $\begingroup$ This is the idea. Nitpick: the solution modulo $x^2+1$ is not unique. Square roots rarely are! $-2x$ works as well as $2x$ :-) $\endgroup$ Sep 8 '21 at 11:22
  • $\begingroup$ @JyrkiLahtonen Good point, I'll add that in. I'll be honest, I wanted to mark this as an abstract duplicate of some abstract lifting-type question, but I've not done so because I thought that lifting , in this case, merited an example to contrast with the OP's approach specifically. I could be wrong, in which case I'll delete. $\endgroup$ Sep 8 '21 at 11:36

Question: "But Is there any other simple way?"

Answer: Here is another approach using formal power series. It gives a solution to your problem for any power $(x^2+1)^{l+1}$ for $l\geq 1$. It is based on the fact that the quotient ring $\mathbb{R}[x]/(x^2+1)^{l+1}$ is a local ring with many units.

The ideal $I:=(x^2+1) \subseteq A:=\mathbb{R}[x]$ is a maximal ideal and the ring $B:=A/I^{l+1}$ is a local ring with maximal ideal $I/I^{l+1}$. Let $T:=\overline{x^2+1}\subseteq A/I^{l+1}$ and consider the element

$$SQ1\text{ }u_l:=1+T+T^2+\cdots +T^l "=" \frac{1}{1-T}.$$

Since $T$ is nilpotent it follows $u_l(1-T)=1$, hence $u$ is a unit in $B$ with inverse $1-T$. There is an element $w_l\in B$ with $w_l^2=u_l$. An explicit element $w_l$ can be constructed using the equation

$$EQ.\text{ }w_l^2:=(a_0+a_1T+\cdots +a_lT^l)^2=u_l $$

and solving for the "variables" $a_i \in k:=\mathbb{R}$. The element $u_l$ has the property that $x^2u_l"="x^2(\frac{1}{-x^2})=-1$. Let $z_l:=2xw_l \in B$ it follows


Hence the element $z_l\in B:=A/I^{l+1}$ solves the equation $p(z_l)=0$ where $p(t):=t^2+4\in A/I^{l+1}[t]$ for any $l\geq 1$.

Example: For $l=1$ you get $w_1:=1+\frac{1}{2}T$. For $l=2$ you get $w_2:=1+\frac{1}{2}T+\frac{3}{8}T^2$. Hence

It follows $z_1:=2xw_1$ has $z_1^2=-4$ and $z_2:=2xw_2$ has $z_2^2=-4$. You get


and $x^2(1+T)=(x^2+1-1)(x^2+1+1)=(x^2+1)^2-1 \cong -1$. Hence


Similarly you get



$$x^2(1+T+T^2)=((x^2+1)-1)((x^2+1)^2+(x^2+1)+1) \cong -1$$

hence $z_2^2 \cong 4x^2(1+T+T^2) \cong -4$.

Hence the problem is solved by the fact you can calculate the square root $w_l:=\sqrt{u_l}\in A/I^{l+1}$ in $SQ1$. There are projection maps

$$\pi_l: A/I^{l+1} \rightarrow A/I^{l}$$

and $\pi_l(z_l)=z_{l-1}$. Hence you get a compatible system of solutions $\hat{z}:=(z_l)_l \in \hat{A}$ in the completion of $A$ wrto $I$: There is an element $\hat{z} \in \hat{A}$ such that


Here $p(t):=t^2+4\in \hat{A}[t]$.

Note: The local ring $(\hat{A}, \hat{\mathfrak{m}})$ has $\mathbb{C}$ as residue field. Hence you have "lifted" a solution of the equation $z^2=4$ in $\mathbb{C}$ to $\hat{A}$. The "lifted" solution gives you a solution in $A/\mathfrak{m}^{l+1}$ for all $l\geq 0$.

  • $\begingroup$ This does not look simpler (to me) but yes is rather more general. Unfortunately this is beyond my grasp right now. $\endgroup$
    – Paramanand Singh
    Sep 10 '21 at 2:24
  • $\begingroup$ @ParamanandSingh - to calculate the square root of $1+T$ and $1+T+T^2$ is an elementary calculation. The approach also allows one to generalize to the formal completion. $\endgroup$
    – hm2020
    Sep 10 '21 at 7:27
  • $\begingroup$ @ParamanandSingh - The equation $EQ$ can be solved in any algebra $A/I^{l+1}$ over the rational humbers, with $u_l=1+x+x^2+ \cdots + x^l$ and $x\in I$. Hence you can take the square root $\sqrt{u_l}$ of any such element. $\endgroup$
    – hm2020
    Sep 10 '21 at 8:47

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