abstract algebra - Remainders of functions

I was given this question:

Consider $\mathbb{R}[x]$ and lets $\mathbb{R}_1[x]=\{a+bx:a,b\in\mathbb{R}\}$. Define the map $\phi:\mathbb{R}[x] \rightarrow \mathbb{R}_1[x]$ by letting $\phi(f(x))$ be the remainder $r(x)$, when $f(x)$ is divided by $x^2+1$. This is well-defined, and in $\mathbb{R}_1[x]$, by the division algorithm in $\mathbb{R}[x]$.

I would like to calculate $\phi(3x^2+4x+7)$ but I am confused as to what is being asked here.

The wording is throwing me off, so I am assuming that I want to take $3x^2+4x+7$ and divide it by $x^2+1$, and the outputted remainder, $r(x) \in \mathbb{R}_1[x]$ is what I am looking for?

How do I approach this type of problem?

Answer

The problem is saying that your map gives you the reminder of some polynomials (with real coefficients) divided by $x^2+1$. So if you take a polynomial $p(x)\in\mathbb{R}[x]$ and divide this polynomial by $(x^2+1)$ you get a new polynomial $q(x)$ such that $$p(x) = (x^2+1)q(x)+ R(x)$$ where $R(x)$ is the reminder of the division. Your function then $$\phi:\mathbb{R}[x]\rightarrow \mathbb{R}_1[x]\\\phi(p(x))\mapsto R(x)$$ As a trivial example, let's take the polynomial $p(x) = x^2+1$, obviously $$\phi(p(x))=\phi(x^2+1)=0$$ so the kernel of this map is all the polynomials of the type $k(x^2+1)$ where $k\in\mathbb{R}$.

To find the value of the map for the polynomial $3x^2+4x+7$, instead of doing long division, you just have to do extrapolate a $(x^2+1)$ factor, for as many times as you can, from that polynomial, mainly: $$3x^2+4x+7 = 3(x^2+1)+4x+4$$ so that $$\frac{3x^2+4x+7}{x^2+1}=\frac{3(x^2+1)+4x+4}{x^2+1}= \\ = \frac{3(x^2+1)}{x^2+1} + \frac{4x+4}{x^2+1} = 4\frac{x+1}{x^2+1}+3$$ clearly the reminder is $3$. So in the end $$\phi(3^2+4x+7) = 3$$

Edit

The fact that your reminders $R(x)$ have to be in $\mathbb{R}_1[x]$ means that you have to do your division until you get something of the form $a+bx$ and not a polynomial of higher degree. For example, if you take some polynomial of fifth degree, call it $$a(x) = a+bx+cx^2+dx^3+ex^4+fx^5$$ then, dividing by $x^2+1$ you only get to the point were $$a(x) = q(x)(x^2+1)+R(x)$$ where $$q(x) = a+bx+cx^2$$ Surely you could divide $q(x)$ again by $x^2+1$ but this would give you, in total, a reminder of degree more than one.