1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.