Thursday 17 July 2014

abstract algebra - 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)?

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...