Example to show that $\mathbb Z^3$ is not a quotient of $\mathbb Z[X]$

Question

I am asked to prove that $\mathbb{Z}^n$ is not a quotient of $\mathbb{Z}[X]$ for any integer $n \geq 3$ by showing that there is no surjective ring homomorphism $\mathbb{Z}[X] \to \mathbb{Z}^n.$

We can just focus on the $n = 3$ case since we can always look at the first 3 entries of the image. If $\phi: \mathbb{Z}[X] \to \mathbb{Z}^3$ is a ring homomorphism, then restricted to each entry, it is still a ring homomorphism $\phi_i: \mathbb{Z}[X] \to \mathbb{Z}, i = 1, 2, 3$. Now, $\phi_i$ must be the identity on $\mathbb{Z}$, and it is completely determined by $\lambda_i := \phi_i(X)$, so it is essentially the evaluation homomorphism $p \mapsto p(\lambda_i)$.

But $p(\lambda_i)$ is the remainder of $p(X)$ modulo $X - \lambda_i,$ so it reminds me of how in $\mathbb{Z},$ there is no integer $n$ satisfying $n \equiv 0 \bmod 3$ and $n \equiv 1 \bmod 6$. However, if the $\lambda_i$s are distinct, then the polynomials $X - \lambda_i$ look "coprime" to me, and I was wondering about whether there is a version of Chinese Remainder Theorem for ideals. Moreover, when looking at the examples, I noticed that (for instance) there is a surjective homomorphism $p \mapsto (p(0), p(1))$ since $(m, n)$ has pre-image $(m-n)X + n$, so why can we not extend it to a third remainder? More generally, how would the homomorphism fail to be surjective for any $\lambda_1, \lambda_2, \lambda_3$?

Answer 1

There is a pretty straightforward way to see why. As you correctly pointed out, a ring homomorphism $\varphi :\mathbb{Z}[X]\rightarrow\mathbb{Z}^3$ is nothing more than an evaluation homomorphism in each component. So there are $\alpha$, $\beta$, $\gamma\in\mathbb{Z}$ such that $\varphi(p)=(p(\alpha), p(\beta), p(\gamma))$ for every $p\in\mathbb{Z}[X]$.

Now, observe that $\varphi$ is surjective if and only if its image contains the elements $(1,0,0), (0,1,0), (0,0,1)$. This means that there are three polynomials $p_1, p_2, p_3 \in\mathbb{Z}[X]$ such that:

-$p_1(\alpha)=p_2(\beta)=p_3(\gamma)=1$;

-$p_1(\beta)=p_1(\gamma)=p_2(\alpha)=p_2(\gamma)=p_3(\alpha)=p_3(\beta)=0$.

What does it tell you about $\alpha, \beta, \gamma$?

Answer 2

Unfortunately, in order for the ideals $(X - a),$ $(X - b),$ and $(X - c)$ to be pairwise comaximal (i.e., in order for $(X - a) + (X - b) = \mathbb Z[x],$ etc.), we would need \begin{align*} b - a &= (X - a) - (X - b) = \pm 1, \\ c - a &= (X - a) - (X - c) = \pm 1, \text{ and} \\ c - b &= (X - b) - (X - c) = \pm 1. \end{align*} Considering that $c - b = (c - a) - (b - a)$ (for example), it cannot be the case that all three of these equations hold simultaneously.

On the other hand, observe that $X - (X - 1) = 1$ so that $(X) + (X - 1) = \mathbb Z[x].$ Consequently, by the Chinese Remainder Theorem, we have that $$\frac{\mathbb Z[x]}{(X) \cap (X - 1)} \cong \frac{\mathbb Z[X]}{(X)} \times \frac{\mathbb Z[X]}{(X - 1)} \cong \mathbb Z \times \mathbb Z.$$

Answer 3

Suppose $p(x) \in \mathbb{Z}[x]$ a polynomial and $deg(p) > 1$.

Let $\lambda_{1} \neq \lambda_{2}$ be integers such that $p(\lambda_{1}) \neq p(\lambda_{2})$, for a certain $p(x) \in \mathbb{Z}[x]$.

Then there are $f_{1}(x), f_{2}(x) \in \mathbb{Z}[x]$ such that, for maximal values of the exponents $m_{1}, m_{2}$ we can write:

$$p(x) = p(\lambda_{1}) + (x-\lambda_{1})^{m_{1}}f_{1}(x),$$

$$p(x) = p(\lambda_{1}) + (x-\lambda_{1})^{m_{1}}[(x-\lambda_{2})^{m_{2}}f_{2}(x) + f_{1}(\lambda_{2})].$$

Using this last equation, we have that:

$$p(\lambda_{2}) - p(\lambda_{1}) = (\lambda_{2} - \lambda_{1})^{m_{1}}f_{1}(\lambda_{2}).$$

Because $deg(p) > 1$ and $p(\lambda_{2}) \neq p(\lambda_{1})$, we have that $m_{1} \geq 1$ and $f_{1}(\lambda_2) \neq 0$ so, we have that $(\lambda_{2} - \lambda_{1}) | (p(\lambda_{2}) - p(\lambda_{1}))$.

Now let us get back to your problem: if there is such a surjective homomorphism $p \mapsto (p(\lambda_{1}), p(\lambda_{2}), p(\lambda_{3}))$ (all $\lambda_{i}$'s are distinct), then there would exist infinite values $(a,b,c) \in \mathbb{Z}^{3}$ that are the image of polynomials of degree at least 2 (since the polynomials of degree $\leq 1$ form a $\mathbb{Z}$-module of rank 2, and $\mathbb{Z}^{3}$ has rank 3). Now, for such polynomials, if $p(\lambda_{i}) \neq p(\lambda_{j})$ we know that:

$$(\lambda_{i} - \lambda_{j})|(p(\lambda_{i}) - p(\lambda_{j})).$$

We can assume that $\lambda_{2} - \lambda_{1} = k > 1$. Therefore, for every $(a, a+k+1, c)$, there must be a linear polynomial $p$ such that $p(\lambda_{1}) = a, p(\lambda_{2}) = a + k + 1$ and $p(\lambda_{3}) = c$, and this cannot be true for all values of $c$. Hence, we reached our contradiction.