Proof that $x \in \Bbb C[x]/\langle p(x)\rangle$ has a square root

Question

I'm looking for a "nice" proof of the following result.

Proposition: Let $p \in \Bbb C[x]$ be such that $p(0) \neq 0$. Prove that there exists a $q \in \Bbb C[x]$ such that $$ q(x)^2 = x \pmod {p(x)}. $$ In other words, $x + \langle p\rangle \in \Bbb C[x]/\langle p\rangle$ is a perfect square.

Here is a proof that I don't find particularly nice.

Let $V = \Bbb C[x]/\langle p\rangle$; note that $V$ is a finite-dimensional vector space. The multiplication operator $T:V \to V$ define by $[Tf](x) = xf(x)$ defines an invertible linear operator. By various means (cf. this post or this post), it can be shown that there exists an operator $S:V \to V$ for which $S = q(T)$ for some $q \in \Bbb C[x]$ and $S^2 = T$. It follows that $[q + \langle p\rangle]^2 = x + \langle p\rangle$.

This proof relies on linear-algebraic machinery, and in particular on Jordan normal form. Is there a nicer approach, perhaps one that is "purely algebraic"?

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Here's a formal extension of Eric's "explicit" construction. With the application of the Chinese remainder theorem and the substitution $y = x-a$, the problem is reduced to that of showing that $y + a$ has a square root modulo $y^n$. We can see that such a square root exists as follows. Define $$ q(y) = b_0 + b_1 y + \cdots + b_{n-1} y^{n-1}. $$ We find that the coefficients of $q^2$ are given by the "Cauchy product", i.e. $$ q(y)^2 = b_0^2 + (2b_0b_1) y + (2b_0b_2 + b_1^2)y^2 + \cdots + (2b_{n-1}b_0 + \cdots)y^{n-1} \pmod{y^n}. $$ Define $a_j$ to be the coefficient of $y^j$ in the above product. We note that for $j \geq 1,$ $$ a_j = 2b_0b_j + f_j(b_0,\dots,b_{j-1}). $$ If we take $b_0 = \sqrt{a} \neq 0$, we then get $a_0 = a$ as desired and find that the system $$ a_1 = 1, \quad a_j = 0 \text{ for } j \geq 2, $$ can be solved recursively by taking $$ b_1 = \frac 1{2b_0}, \quad b_j = -\frac{f_j(b_0,\dots,b_{j-1})}{2b_0} \text{ for } j \geq 2. $$ Thus, there indeed exists a polynomial $q(y)$ for which $[q(y)]^2 = x + a$, which was what we wanted.

Answer 1

Well, this isn't really any different from using Jordan normal form, but here's how I'd think about it. By the Chinese remainder theorem, the quotient $\mathbb{C}[x]/(p(x))$ splits as a product of $\mathbb{C}[x]$-algebras of the form $\mathbb{C}[x]/((x-a)^n)$, where the $a$'s are the roots of $p$ and the $n$'s are their multiplicities. So, it suffices to show that $x$ is a square in $\mathbb{C}[x]/((x-a)^n)$ if $a\neq 0$. Or, setting $y=x-a$, it sufffices to show that $y+a$ is a square in $\mathbb{C}[y]/(y^n)$. This follows immediately from the fact that $a$ is a square in $\mathbb{C}$ and Hensel's lemma. Explicitly, you can construct a square root of $y+a$ by starting with $b\in\mathbb{C}$ that is a square root of $a$, then taking $\frac{y}{2b}+b$ to get a linear term of $y$ when you square it, then adding a quadratic term to make the quadratic term of the square vanish, and so on up to the $y^{n-1}$ term.