Is $x^2 + 1$ irreducible polynomial in $\Bbb R[x,y]$?

Question

Is $x^2 + 1$ is irreducible polynomial in $\Bbb R[x,y]$?

Actually I recently got to know that $\Bbb R[x,y] \over \langle x^2 +1\rangle$ is isomorphic to $\Bbb R[i][y]$ which is a PID not a field. So $x^2 + 1$ is certainly not irreducible.

Can anyone please help me to find two polynomial in $\Bbb R[x,y]$ whose multiplication is $x^2 + 1$?

Answer 1

$\;x^2+1\;$ indeed is irreducible in $\;\Bbb R[x,y]\;$ , yet the ideal $\;\langle x^2+1\rangle\;$ is not maximal in that ring since, for example

$$\langle x^2+1\rangle\lneq\langle x^2+1\,,\,\,y\rangle\le\Bbb R[x,y]$$

Thus, the quotient ring can't indeed be a field...

Answer 2

The quotient ${\Bbb R}[x,y]/\langle x^2+1\rangle \cong {\Bbb C}[y]$ is a domain (but not a field). So the ideal $\langle x^2+1 \rangle$ is prime (but not maximal) and $x^2 + 1$ is irreducible.