Zero of $x^2 + 1$ in $\mathbb{Q}[x] / \langle (x^2 + 1)\rangle$

Question

Let $f(x)$ = $x^2 + 1$ be a polynomial in $\mathbb{Q}[x]$. Clearly, $f$ has no zeroes in this ring. Construct an extension field $E$ in which $f$ has a zero.

The answer is $\mathbb{Q}[x] / \langle (x^2 + 1)\rangle$. But I don't really quite see what the zero is, or how to construct a zero.

Really, any help on this would be tremendously helpful. Thank you.

Answer 1

We want to check that there is an element $\alpha = p(x) + \langle x^2 + 1\rangle\in E$ such that $\alpha^2 + (1 + \langle x^2 + 1\rangle) = 0 + \langle x^2 + 1\rangle$. Let $\alpha = x + \langle x^2 + 1\rangle$. Then $$ \alpha^2 + (1 + \langle x^2 + 1\rangle) = (x + \langle x^2 + 1\rangle)^2 + (1 + \langle x^2 + 1\rangle) = (x^2 + \langle x^2 + 1\rangle) + (1 + \langle x^2 + 1\rangle) = (x^2 + 1) + \langle x^2 + 1\rangle = 0 + \langle x^2 + 1\rangle, $$ as desired.

Answer 2

Check for :

$x+\langle x^2+1\rangle$

Answer 3

Consider the image of x, under the natural map from Q[X] to Q[X]/ . That is your root of the polynomial.