# Are these rings fields?

Are the following rings fields?

1) $\Bbb Q[x] /\langle x^2+1\rangle$

Since a polynomial ring taking values on any field is a E.D, and hence a P.I.D, this is a field iff the ideal is prime or maximal.

Any irreducible in this quotient ring is a maximal ideal, and $x^2+1$ is an irreducible polynomial in the quotient ring, since we don't have algebraic closure, hence the quotient ring is a field.

2) $\Bbb F_2[x] / \langle x^2 +1 \rangle$

Not sure how to show if $x^2+1$ is irreducible here, I have a feeling it isn't but no way to expand on that. Some comments, I do know that $\Bbb F_2$ means I am taking values from $\{0,1\}$ and hence $-1=1 \pmod 2$

3) $\Bbb Q[x] / \langle x^4 + 6x^3 + 9x + 6\rangle$

Here I just need to see if $x^4+6x^3+9x+6$ is irreducible.

What I did was a bit strange:

$$x^4+6x^3+9x+6$$ $$=(x+1)x^3+5x^3+9x+6$$ $$=x^3((x+1)+5)+9x+6$$ $$=x(x^2((x+1)+5)+9)+6$$

Which gives us the root $x=-6$ and hence this will be generated by $(x+6)$, so this is not maximal, hence this quotient ring is not a field. This feels sketchy, since perhaps that root is in $\Bbb Q$ but the other ones aren't.

Is my logic correct in 1)&3), how do I do 2)?

• Hint for (2): does $x^2+1$ have roots in $\mathbb F_2$ ? – user228113 Jun 21 '15 at 6:07
• @G.Sassatelli Well I suppose it equals $(x+1)(x-1)$ since $1=-1$ mod, so then I could say $(x+1)$ or $(x-1)$ generates this, so those are maximal, but this is not – homomorphism Jun 21 '15 at 6:09
• You could say that $\langle x^2+1\rangle\mathbb F_2[x]$ is not a prime ideal because $x^2+1$ is not irreducible in $\mathbb F_2[x]$ – user228113 Jun 21 '15 at 6:12
• @G.Sassatelli Very good! – homomorphism Jun 21 '15 at 6:16
• For the last one, one usually uses a test which is known as Eisenstein's criterion (it is actually irreducible). Have you seen this test yet? – Adam Hughes Jun 21 '15 at 6:22

It is incorrect. $-6$ is not a root of $x^4+6x^3+9x+6$. Let's look for rational root of that polynomial ${p\over q}$.

We have $q|1$ and $p|6$. (Do you see why?)

So the only rational roots can be $\pm 6$. None of them is and to prove the irreducibility we need more because this quartic could for instance decompose as a product of two irreducible quadratics.

Eisenstein criterion works nicely with $p=3$ in this case

• Unfortunately just because it doesn't have rational roots, doesn't mean it isn't irreducible. Consider $(x^2-2)^2$. – CPM Jun 21 '15 at 6:27
• I was editing while you were commenting – marwalix Jun 21 '15 at 6:29
• Okay, nice. It is a common mistake, so just wanted to make sure to point that out. – CPM Jun 21 '15 at 6:30

Looks good to me. For 2), remember $(x+1)^2=x^2+2x+1=x^2+1$. In general, in a field of characteristic $p$, you get the so called "freshman's dream" $(x+y)^p=x^p+y^p$.

Edit: For 3) I think -6 is not a root in the polynomial you typed, there was an error somewhere. I think this polynomial is actually irreducible too over $\mathbb{Q}$. It apparently factors like this over $\mathbb{C}$: $$(x+0.560333) (x+6.20842) (x-(0.384378+1.25578 i)) (x-(0.384378-1.25578 i))$$

As noted in the comments, if you know Eisenstein's criterion, you are set with $p=3$ to show it is irreducible. Without that, you might have more work.

• I am especially concerned about my sketchy logic for $3)$, why would finding that one root give me the confidence to expect the entire polynomial to factorise nicely? – homomorphism Jun 21 '15 at 6:11