Polynomial rings- multiplicative inverse

Question

I need to solve the following question in ring theory.

Show that $(Q[x])/\langle{x^2+x+4}\rangle$ is a field.

To show that $(Q[x])/\langle{x^2+x+4}\rangle$ is a field, the only thing I need to do is to show that the polynomial $x^2+x+4$ is irreducible. I was able to solve that problem.But the other part which is given below is bit difficult to proceed. I will be happy, if you could kindly help me on this problem.

Let $y=x+\langle{x^2+x+4}\rangle $ be an element in this field.

Show that (i) $y^2+y+4=0$ is in this field.

(ii) ${-\frac{1}{4}(1+x)+N}$ is the multiplicative inverse of ${x+N}$ in $(Q[x])/N$ where $N=\langle{x^2+x+4}\rangle $.

Answer 1

By definition of operations in quotient ring: putting $\;N:=\langle x^2+x+4\rangle\;$, we get

$$y^2+y+4= \left(x+N\right)^2+\left(x+N\right)+(4+N):=x^2+N+x+N+4+N:=$$

$$=(x^2+x+4)+N=N\iff x^2+x+4\in N$$

and since the last relation is obviously true we're done.

Now

$$(x+N)\left(-\frac14(x+1)+N\right):=-\frac14x(x+1)+N=-\frac14x^2-\frac14x+N\;\;(**)$$

But since

$$x^2+x+4=\overline 0= N\implies -x^2-x=\overline 4\implies-\frac14x^2-\frac14x=\overline 1\implies $$

$$(**) = 1+N$$

Answer 2

Hint $\, $ mod $\,\ x\,f- c\ $ we have $\ x\, f \equiv c\ $ so $\ x\, f/c \equiv 1\,\Rightarrow\, x^{-1}\! \equiv f/c,\, $ when $\ c\neq 0$

Yours is case $\, x\,f - c = x(x\!+\!1) -(-4)\ $ so $\ f = x\!+\!1,\ c = -4,\ $ so $\ x^{-1}\! \equiv f/c \equiv -(x\!+\!1)/4$

Remark $\ $ When developing intuition about quotient rings it is very helpful to become proficient translating back-and-forth between coset notation and modular arithmetic notation (as above), since the latter allows us to more efficiently reuse our well-honed arithmetical intuition. Once you become proficient at this it will become subconscious.