Prove that $x^2+I=-1+I$

Question

I have given an ideal $I=\{(x^2+1)\cdot p(x)|p(x)\in \mathbb{R}[x]\}$ which is an ideal of $\mathbb{R}[x]$. I have to prove that $x^2+I=-1+I$ holds in $\mathbb{R}[x]/I$.

I think, that I is also a principal ideal and in the university lecture we discussed the following:

Let $K[x]$ be a ring and $(P(x))$ a principal ideal, then $K[x]/P(x)$ is a quotient ring and the following holds:

$A(x)\equiv B(x) \mod P(x) \Leftrightarrow P(x)|A(x)-B(x)$ as well as $A(x)\equiv B(x) \land C(x)\equiv D(x)\mod P(x)\implies A(x)+(C(x)\equiv B(x)+D(x)\mod P(x)$

I used this to derive:

$x^2+1|x^2-(-1)\Leftrightarrow x^2\equiv -1$ and also $x^2\equiv -1 \land I\equiv I\implies x^2+I\equiv -1+I$ and because $\mathbb{R}[x]/I$ is a quotient ring, $x^2+I=-1+I$ hold as well.

Can anybody check whether this reasing is correct?

Answer 1

To begin with, $I = (X^2+1)\mathbb{R}[X]$ so it is a principal ideal by definition.

As a remark, all ideals of $\mathbb{R}[X]$ are principal (consider the monic polynomial $P$ of smallest degree in the ideal, which is unique by construction, and use the Euclidean division to prove that the ideal is equal to $P\mathbb{R}[X]$). We say that $\mathbb{R}[X]$ is a principal ideal domain (aka a PID).

Your question is whether $X^2+I = -1 + I$ holds in $\mathbb{R}[X]/I$. Be careful with the fact that with conventional notations, $X^2+I$ usually denotes a set, that of all polynomials of $I$ plus $X^2$. If the question is to be understood as the fact that this holds for all polynomials in $I$, this is equivalent to the fact that $X^2=-1$ in $\mathbb{R}[X]/I$ (take $0\in I$ for instance).

And this equality follows directly from the definition of a quotient ring: $A = B$ in $\mathbb{R}[X]/I$ if and only if $A-B\in I$. In our case, since $I = (X^2+1)\mathbb{R}[X]$, the condition becomes $(X^2+1)| A-B$, which is trivially true with $A=X^2$ and $B=-1$.

As a remark, $\mathbb{R}[X]/(X^2+1)$ formally defines complex numbers.

Regarding your reasoning now,

1. the second property you list is wrong, take for example $A=1, B=X, C=X+1, D=1$ (where $B\wedge C=1)$,

2. $X^2\equiv-1\wedge I\equiv I$ is wrong as well: $-1\wedge I = -1$ (and $X^2\equiv I$ does not hold for all polynomials $I$ of the ideal).

Answer 2

To answer your question in short: Yes, your reasoning is correct.

If $I \subset R$ is an ideal and $r_1,r_2 \in R$ are elements of $R$, we have $$ r_1+I = r_2+I \iff r_1-r_2 \in I $$ In this case, it means we have $X^2+I = -1+I \iff X^2+1 \in I$ which is obviously true.