Prove $\langle x^2+1\rangle$ is maximal ideal of $\Bbb{Q}[x]$

Question

I am self-studying some content on ring theory.

I came across this question in which I am supposed to prove that $\langle x^2+1\rangle $ is maximal ideal of $\Bbb{Q}[x]$.

Let us define $\phi : \Bbb{Q}[x]\rightarrow \Bbb Q[i]$

as $\phi(f(x)) = f(i)$

Here $\Bbb{Q}[x]$ is a polynomial ring iwith coefficients from $\Bbb{Q}$ in variable $x$ and $\Bbb{Q}[i]$ is a poynomial ring with variables from $\Bbb{Q}$ in variable $x$.

We know that the latter is a ring.

I have proven that $\phi$ is an onto ring homomomorphism with Kernel = $\langle x^2+1\rangle$.

So by the Fundamental theorem of Homomorphism, we have that

$\frac{\Bbb{Q}[x]}{\langle x^2+1\rangle}\simeq \Bbb Q[i]$.

Is this working correct? Now to prove my desired claim, should I prove that $\Bbb{Q}[i]$ is a field?.

If I am able to do that,them I think I will be done.

So, please help!

Answer 1

Well the easiest way is to note that $X^2+1$ is an irreducible polynomial in $\Bbb Q[X]$. Now since $\Bbb Q[X]$ is a P.I.D, the principle ideal $\langle X^2+1 \rangle$ is maximal(in a P.I.D $\langle a \rangle$ is maxmial $\iff$ $a$ is an irreducible element)

Answer 2

Hint: $\Bbb Q[i]$ is the ring extension of $\Bbb Q$ via the imaginary unit. We have $\Bbb Q[i] = \{a + bi\mid a,b\in\Bbb Q\}$.

This ring extension is also a field extension of $\Bbb Q$. For this, you basically need to show that each nonzero element is invertible. You get the formula by considering

$(a+bi)(c+di) = (ac - bd) = (ad+bc)i$ which you need to set $1 = 1 + 0i$ and compare coefficients.

Answer 3

In fact the logic goes the other direction: you prove that $\Bbb Q[i]$ is a field by showing that $x^2+1$ generates a maximal ideal of $\Bbb Q[x]$.

To show the latter, use the fact that $\Bbb Q[x]$ is a Euclidean domain, hence a PID.

It is a general result that if $f$ is an irreducible element of a PID $A$, then the quotient $A/fA$ is a field, because Bezout's identity gives the inverse of any nonzero element.