I need to determine if the following ring is a field.

$$\mathbb Q[X]/(X^4-2X^2+8X+1)$$ I assume I need to show that it is commutative, has a multiplicative identity and that every non zero element is has a multiplicative inverse, but I am unsure how to go about this.

  • $\begingroup$ Does $2^2=X^2$? In any case, what you need is that the polynomial be irreducible over $\Bbb Q$. $\endgroup$ – Lord Shark the Unknown Mar 15 '18 at 10:06
  • $\begingroup$ Is this the case with all similar type questions too? Show that the polynomial is irreducible over $\mathbb Q$ or any other ring like $\mathbb Z_3$ ? $\endgroup$ – jdminer Mar 15 '18 at 10:08
  • $\begingroup$ @jdminer Well what do you know about taking a quotient by a maximal ideal? $\endgroup$ – Alex Clark Mar 15 '18 at 10:09

It suffices to show that the polynomial $f(x)=x^4-2x^2+8x+1$ is irreducible.

(Because $\mathbb{Q}$ is a field, hence $\mathbb{Q}[x]$ is a P.I.D. and hence every prime ideal is maximal...)

First notice that if the polynomial has roots in $\mathbb{Q}$ it would be $1,-1$(Rational root theorem). So the polynomial $f(x)$ have no rational roots.

Now it is enough to check if it's irreducible in $\mathbb{Z}[x]$ (Gauss lemma).

Hence if it's not irreducible it's must be the product of two polynomials of $\mathbb{Z}[x]$ with degree 2.

Assume that this is true, and take

$x^4-2x^2+8x+1= (a_1x^2+b_1x+c_1)(a_2x^2+b_2x+c_2)$

Solve the system and you'll have a contradiction!

There are a few faster ways to do this, but this is a standard way!


$x^4-2x^2+8x+1$ is irreducible mod $3$ and so is irreducible over $\mathbb Q$.

  • $\begingroup$ This is nice: I would have used Eisenstein. $\endgroup$ – Lubin Mar 15 '18 at 16:45

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