Are $\mathbb{R}[X] / (X^2 +1)$ and $\mathbb{C}$ homeomorphic?

The fields $\mathbb{R}[x] / (x^2 +1)$ and $\mathbb{C}$ are isomorphic as fields, but I am trying to see if they are homeomorphic as well.

$\mathbb{C}$ is given its standard topology, and we can define a metric on $\mathbb{R}[x]$ by $$ d \left( \sum_{j = 0}^a a_j x^j ~, \sum_{j = 0}^b b_j x^j \right) := \left( \sum_{j = 0}^{\max\{a,b\}} (a_j - b_j)^2 \right)^{1/2} $$ which induces a topology. $\mathbb{R}[x] / (x^2 +1)$ is then given the quotient topology.

Any polynomial $[f] \in \mathbb{R}[x] / (x^2 +1)$ can be written as $[f] = a [1] + b [x]$ with $a,b \in\mathbb{R}$. A field isomorphism can then be given by the map $\psi : \mathbb{R}[x] / (x^2 +1) \rightarrow \mathbb{C}$ with $$ \psi([1]) = 1 ~~~,~~~ \psi([x]) = i $$ We then have that $\psi$ combined with the quotient map $\pi: \mathbb{R}[x] \rightarrow \mathbb{R}[x] / (x^2 +1)$ gives $$ \psi \circ \pi \left( \sum_{k = 0}^a a_j x^k \right) = \psi \left( \sum_{k = 0}^a a_k [x]^k \right) = \sum_{k = 0}^a a_k i^k = \operatorname{eval}_i \left( \sum_{k = 0}^a a_j x^k \right) $$ I checked that for any convergent sequence $f_n \rightarrow f$ we have $\lim_{n \rightarrow \infty} \operatorname{eval}_i(f_n) = \operatorname{eval}_i(f) $ and since $\mathbb{R}[x]$ and $\mathbb{C}$ are metric spaces, this shows that $\operatorname{eval}_i = \psi \circ \pi$ is contiuous. By the universal property of quotient maps, this also shows that $\psi$ is continuous.

But I am kind of stuck on proving the continuity of $\psi^{-1}$ and would be glad for any ideas.

Another thing I am questioning is, whether the topology I chose on $\mathbb{R}[x]$ is the most natural one you could choose. The second possibility I could think of is to look at $\mathcal{C}(\mathbb{C},\mathbb{C})$ with the compact-open topology and consider $\mathbb{R}[x]$ as a subspace if this (continuity of $\operatorname{eval}_i$ would easily follow in this case). It would be interesting to know if they are homeomorphic or not.

Update

As Paul Frost poited out (and showed), $\psi$ is not continuous with the assumptions I made. I also went back to check my calculations on the continuity of $\operatorname{eval}_i$ and indeed found a mistake there.

He also showed that $\mathbb{R}[x]$ equipped with the topology induced by the $\Vert \cdot \Vert_1$ norm does make $\psi$ continuous. Since the proof by paul blart math cop for the continuity of $\psi^{-1}$ remains correct under these circumstances, $\psi$ is shown to be a homeomorphism in this case. So this anwsers the first part of my question.

Solution 1:

Let $ \varphi: \mathbb{C} \to \mathbb{R}[t] $ via $ \varphi(1) = 1, \varphi(i) = t $ an $ \mathbb{R} $ vector space map. Then $ \varphi $ is an isometry onto its image and is therefore continuous. Also, $ \psi^{-1} = \pi \circ \varphi $. This is easily verified on $ \{1, i\} $ and both sides of the equation are $ \mathbb{R} $-linear, so they are equal. Thus, $ \psi^{-1} $ is continuous so $ \psi $ is a homeomorphism.

Solution 2:

With the topology induced by your metric $d$ the map $\psi$ is NOT contimuous.

Let us first analyze $\psi$ algebraically. I suggest to introduce this map as follows. By the universal property of the polynomial ring there exists a unique ring homomorphism $\Psi : \mathbb{R}[x] \to \mathbb{C}$ such that $\Psi(x) = i$. Thus explicitly

$$\Psi(\Sigma_{j=0}^r a_j x^j) = \Sigma_{j=0}^r a_j i^j .$$

We have $\Psi(x^2 + 1) = 0$. Hence the ideal generated by $x^2+1$ is mapped to $0$ and $\Psi$ induces a unique ring homomorphism $\psi : \mathbb{R}[x]/(x^2+1) \to \mathbb{C}$ such that $\psi \circ \pi = \Psi$. Obviously this is your isomorphism of fields.

Let us now ask the following general question:

If $\mathbb{R}[x]$ is endowed with a topology $\mathfrak{T}$, then $\mathbb{R}[x]/(x^2+1)$ inherits a quotient topology. Under what conditions on $\mathfrak{T}$ is $\psi$ a homeomorphism?

We only consider the case that $(\mathbb{R}[x],\mathfrak{T})$ is a Hausdorff topological vector space. Then the map $\varphi : \mathbb{C} \to \mathbb{R}[x]$ as defined by user571438 is continuous. This is true because the polynomials of degree $\le 1$ form a Hausdorff two-dimensional linear subspace. It is known that each such linear space is topologically isomorphic to $\mathbb{R}^2$ with its standard topology. The map $\pi \circ \varphi$ is the inverse for $\psi$. We conclude

(1) $\psi$ is a homeomorphism if and only if $\psi$ is continuous.

Moreover we have

(2) $\psi$ is continuous if and only if $\Psi$ is continuous.

Our above question is therefore equivalent to:

Under what conditions on $\mathfrak{T}$ is $\Psi$ continuous?

Your metric $d$ is induced by the norm $\lVert \Sigma_{j=0}^r a_j x^j \rVert_2 = (\Sigma_{j=0}^r a_j^2)^{1/2}$.

Define $p_n(x) = \frac{1}{n} \Sigma_{j=0}^{n-1}x^{4j} \in \mathbb{R}[x]$. Then $\lVert p_n(x) \rVert_2 = \frac{1}{n} n^{1/2} = n^{-1/2}$ so that $p_n(x) \to 0$.

The sequence $\Psi(p_n(x)) = \frac{1}{n} n = 1$ does not converge to $\Psi(0) = 0$ and this means that $\Psi$ is not continuous.

In other words, the topology induced by $d$ on $\mathbb{R}[x]/(x^2+1)$ cannot be Hausdorff (if it were, $\mathbb{R}[x]/(x^2+1)$ would be topologically isomorphic to $\mathbb{R}^2$ and $\psi$ would trivially be continuous).

Let us now consider the norm $\lVert \Sigma_{j=0}^r a_j x^j \rVert_1 = \Sigma_{j=0}^r \lvert a_j \rvert$. Then

$$\lvert \Psi(\Sigma_{j=0}^r a_j x^j) \rvert = \lvert \Sigma_{j=0}^r a_j i^j \rvert \le \Sigma_{j=0}^r \lvert a_j \rvert = \lVert \Sigma_{j=0}^r a_j x^j \rVert_1 .$$

This means that $\Psi$ is continuous with respect to the topology induced by $\lVert . \rVert_1$.

Remark: For all polynomials $p(x)$ we have $\lVert p(x) \rVert_2 \le \lVert p(x) \rVert_1$. This means that $\lVert . \rVert_1$ is stronger than $\lVert . \rVert_2$. In other words, the topology induced by $\lVert . \rVert_1$ is finer than that induced by $\lVert . \rVert_2$. As we have seen above, the two norms are not equivalent because $\Psi$ is continuous with respect to $\lVert . \rVert_1$, but not with respect to $\lVert . \rVert_2$. It can also be seen directly by considering $q_n(x) = \Sigma_{j=0}^{n-1} x^j$. We have $\lVert q_n(x) \rVert_1 = n$ and $\lVert q_n(x) \rVert_2 = n^{1/2}$, i.e. $\lVert q_n(x) \rVert_1/\lVert q_n(x) \rVert_2 = n^{1/2} \to \infty$.