Is the set $I$ a maximal ideal?

Question

If $I = \{f(x) ∈ Q[x] \ | \ \text{f(x) is divisible by} \ x^2 - 1 \}$ is an ideal.

Is $I$ a maximal ideal?

I know the definition of a maximal ideal, but I am struggling to understand how to prove it.

An ideal of a ring R is maximal if and only if

i) I ⊄ R; and

ii) there does not exist an ideal of R, for example J, such that I⊄J⊄R

If anyone could show me how this works with this example it would be a great help!

Also, the next step is to show if the quotient ring Q[x]/I is an integral domain. Any tips or hints for this part will also be appreciated.

Thanks!

Answer 1

A useful characterisation of maximal ideals is the following: the ideal $I$ of $A$ is maximal if and only if $A/I$ is a field. In your case, the polynomial $x^2-1 = (x-1)(x+1)$ is not irreducible, so the ideal it generates is not even prime, that is, $\mathbb{Q}[X]/I$ is not an integral domain. (basicaly because $x-1\neq 0$ and $x+1\neq 0$ in $\mathbb{Q}[X]/I$, but their product is $0$).

Answer 2

$I$ is not a maximal ideal, since it is not even a prime ideal : set $a=(X + 1)$ and$b = (X - 1)$. You have $a \times b = X^2 - 1 \in I$ yet neither $a$ nor $b$ is in $I$.

In general (edit, credit to @Maxime Cazaux), to prove that an ideal $I$ is maximal, one shows that the quotient $R / I$ is a field.

Answer 3

Well, it would be a maximal ideal if the polynomial would be irreducible. But it is not, it decomposes into $x^2-1=(x-1)(x+1)$. It follows that your ideal $I$ is properly contained in the ideal $I_1$ generated by $x-1$ and also in the ideal $I_2$ generated by $x+1$. So $I$ is not maximal. But the ideals $I_1,I_2$ are maximal.

Answer 4

Note that $\mathbb{Q}[x]$ is Pricipal Ideal Domain.

There is a result that says in a PID maximal ideals are exactly the prime ideals.

Here the polynomial $x^2-1=(x-1)(x+1)$ is not irreducible in $\mathbb{Q}[x]$.

So the ideal $I$ is not a prime ideal.

So by the above result, $I$ can't be a maximal ideal.