Is $\mathbb Q$ a quotient of $\mathbb R[X]$?

Question

Is there some ideal $I \subseteq \mathbb R[X]$ such that $\mathbb R[X]/I \cong \mathbb Q$?

$I$ is clearly not a principal ideal.

Answer 1

If such an ideal exists, consider the ring homomorphism obtained by composing the ring homomorphisms

$$\mathbb R \rightarrow \mathbb R[X] \rightarrow \mathbb{R}[X]/I \xrightarrow{\cong} \mathbb Q$$

What can you say about the kernel of this ring homomorphism?

Answer 2

If $R$ is a ring, then $R/I$ is a field $\iff I$ is maximal$^\dagger$. The maximal ideals in any ring of polynomials over a field are precisely those of the form $\langle f \rangle$, where $f$ is an irreducible polynomial$^\ddagger$. What are the irreducible polynomials in $\mathbb{R}[x]$? They are the linear polynomials (in which case the quotient returns $\mathbb{R}$) as well as the quadratics of positive discriminant.

Let's suppose $I = \langle f \rangle$, where $f$ is such a quadratic with $\alpha$ as a root. You can show that $\mathbb{R}[x]/I$ is necessarily isomorphic to $\mathbb{C}$ by applying the isomorphism theorem to the evaluation homomorphism $\text{ev}_\alpha: \mathbb{R}[x] \rightarrow \mathbb{C}$ defined where $g(x) \mapsto g(\alpha)$. You can use a similar approach to show that modding out by a linear polynomial gives $\mathbb{R}$.

In summary, $\mathbb{R}$ and $\mathbb{C}$ are the only possible fields obtained by a quotient of $\mathbb{R}[x]$.

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$^\dagger$ See here for discussion.

$^\ddagger$ The reason these are principally generated as such is because a Euclidean algorithm exists thanks to polynomial long division, wherein the ideal generated by a given set is also principally generated by the "greatest common divisor" of the elements in that set (cf. Euclidean domain).

Answer 3

It could just be me but I feel the existing answers work too hard.

Suppose $f$ is such an isomorphism.

In $\mathbb R[x]/I$, the element $\sqrt 2 +I$ squares to $2+ I$. Since $1+I$ maps to $1\in \mathbb Q$, you have that $f(\sqrt 2+I)$ is a square root of $2$ in $\mathbb Q$. See the problem?

Answer 4

$\Bbb R[X]/I$ is necessarily a vector space over $\Bbb R$, so it is either $\{0\}$ or it is uncountable. It cannot be isomorphic to the nonzero countable $\Bbb Q$.