Show that following quotient ring is field: $R[x]/(x^2+1)$

Question

I need to show that following ring is field.$$R[x]/(x^2+1)$$ I know $(x^2+1)$ is an ideal the unity of the quotient ring is $1+(x^2+1)$. Hence I need to show that for any $f(x)\in R[x]$ I can find $g(x)\in R[x]$ such that $$\{f(x)+(x^2+1)\}\{g(x)+(x^2+1) \}=1+(x^2+1)$$ Expanding on left gives us $$f(x)g(x)+(x^2+1)$$ Hence I need to show that there is a $g(x)$ such that $f(x)g(x)-1\in (x^2+1)$.

Answer 1

Since $x^2+1\nmid f(x)$ and $x^2+1$ is irreducible (in $\mathbb{R}[x]$), $x^2+1$ and $f(x)$ are coprime in $\mathbb{R}[x]$ and therefore thre are $g(x),h(x)\in\mathbb{R}[x]$ such that $f(x)g(x)+(x^2+1)h(x)=1$. So, $f(x)g(x)\in1+(x^2+1)$.

Answer 2

Another hint would be: You're adjoining an element $x$ to the field of real numbers, and by modding out with $(x^2+1)$, you're declaring that this $x$ satisfies $x^2 + 1 = 0$, or $x^2=-1$. Do you know any numbers that satisfy this equation?

Answer 3

A different view would be to consider the ring substitution homomorphism $\phi:{\Bbb R}[x]\rightarrow {\Bbb C}$ with $\phi(f)=f(i)$, where $i$ is a zero of $x^2+1$. This mapping is surjective, since ${\Bbb C} = {\Bbb R}[i] = {\Bbb R}(i)$ (ring adjoint equals field adjoint here). Thus by the Homomorphism theorem, ${\Bbb R}[x]/\ker \phi$ is isomorphic to ${\Bbb C}$. The kernel is a principal ideal generated by the minimal polynomial $x^2+1$ of $i$ over ${\Bbb R}$. Qed.

Answer 4

1. If $A$ is any commutative ring and $P$ is a maximal ideal in $A$, then $A/P$ is a field.

2. If $A$ is a domain and $a$ is an irreducible element in $A$, then $(a)$ is a maximal ideal of $A$.

3. The polynomial $x^2+1$ is irreducible in the domain $\mathbb{R}[x]$.

Once you have proved the three facts above, you're done.