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Let $\mathcal P$ be the associative algebra consisting of real polynomials on the variable $x$. Set $\mathcal C$ to be the ideal of $\mathcal P $ generated by $x^2+1 $.

Why does $\mathcal C$ consist of polynomials of the form $f(x)(x^2+1)g(x)$? (And are $f,g$ in $\mathcal C$ or $\mathcal P$?)

I would denote $\mathcal C $ by $\mathbb R[x^2+1]$, the algebra of polynomials in the variable $x^2+1$.

I would the denote the factor space $\mathcal P/\mathcal C$ by $\mathbb R[x]/\mathbb R[x^2+1]$, all polynomials $p$ satisfying $\frac{\partial p}{\partial (x^2+1)}=0$.

Supplement.

Put $\mathcal F = \mathbb R[x]/(x^2+1)\mathbb R[x] $. We show that the elements of $\mathcal F$ are of the form $p+ (x^2+1)\mathbb R[x] \ ( \mathrm{deg}(p)\leq 1)$.

Consider an arbitrary polynomial $$f = a_n x^n +\ldots+ a_0 \in \mathbb R[x].$$ Writing $$f = a_nx^{n-2}(x^2 +1-1) + a_{n-1}x^{n-3}(x^2 +1-1) + \ldots ,$$ we see that in $\mathcal F$ the polynomial $f$ has degree less then or equal to $1$.

Mussé Redi
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$\mathcal{C}$ consists of polynomials of the form $(x^2+1)f(x)$ where $f(x)$ is another real polynomial. (You do not need both $f(x)$ and $g(x)$ in what you wrote: since multiplication is commutative here, you can just put them on the same side and multiply them into a single polynomial $h(x)$.)

For any commutative ring $R$ and element $a\in R$, the set $aR=\{ar\mid r\in R\}$ is the smallest two-sided ideal containing $a$.

The ideal $\langle x^2+1 \rangle\lhd \Bbb R[x]$ is not the same thing as the algebra of polynomials in the variable $x^2+1$. For example, $1+(x^2+1)$ is such a polynomial, but it is not in the ideal generated by $x^2+1$.

The description using partial derivatives is also not correct. If the derivative of $p$ with respect to $x^+1$ is zero, then $p$ only differs from $x^2+1$ by a constant. Mod $(x^2+1)$ these elements do form one equivalence class in the quotient, but there are still others which are not just constant, namely $x+(x^2+1)$.

It looks like you might be a bit off in your picture of how quotient rings work. If so, you might consider reading questions about quotient rings here, perhaps starting with this one: Calculations in quotient rings.

The ordinary notation for the quotient ring would be $\Bbb R[x]/(x^2+1)$ or $\Bbb R[x]/\langle x^2+1\rangle$, or even $\Bbb R[x]/(x^2+1)\Bbb R[x]$. It's true that $\Bbb R[x^2+1]$ is a perfectly fine notation for the polynomial ring where $x^2+1$ is the variable, but as we discovered, it is not the same thing as the ideal generated by $x^2+1$.

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rschwieb
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  • What kind of cosets do we find in $\mathbb R[x] / (x^2+1)\mathbb R[x]$? I know that if we take, for instance, the quotient group $G=\mathbf Z/3\mathbf Z$, we can denote it by $G={a\mathbf Z\ |\ a \in {1,2,3}\ }$. – Mussé Redi Jul 02 '14 at 16:31
  • @MusséRedi Just as 1,2 and 3 are representatives of cosets in that case, the polynomials of degree 1 or less can be used as representatives for this quotient ring. So if I partially list some, you would have $1+I,2+I,-1+I,\ldots, x+I, 1+x+I, 2x+I$ etc where $I=(x^2+1)$ – rschwieb Jul 02 '14 at 16:35
  • So the notation $\mathbb R[x]/(x^2+1)\mathbb R[x] = {p\mathbb R[x] |\ \mathrm{deg}(p) \leq 1\ } $ is correct? What if we consider $\mathbb R [x]/(x^2+x)\mathbb R[x]$, does the same expression then hold? – Mussé Redi Jul 02 '14 at 16:49
  • @MusséRedi no, the equality you wrote is not correct. It is not multiples of those elements, those elements are additive representatives. You could write ${p+I \mid deg(p)\leq 1}$ where I is the ideal generated by $x^2+1$. – rschwieb Jul 02 '14 at 18:20