# Need hints on the following algebra problems.

I've been looking at these for over an hour and I don't understand how to do them. Any hints would be greatly appreciated.

Let $p(x) = x^3 + x + 1$ and $F = Z_3[x]/\langle p(x)\rangle$. Factor $p(x)$ in $F(x)$. Does it factor into linear terms?

Here I am just supposed to factor it out only if it has a root in F(x). True?

List a complete set of representatives of $F[x]/\langle p(x)\rangle$ where $F = Z_2$ and $p(x) = x^4 + x^2 + 1$.

Any quick way to do this? Am I supposed to write down the multiplication table?

Find the multiplicative inverse of $x^2 + 1$ in $Q[x]/\langle x^4 - 2\rangle$.

Here I let $x^4 \cong 2 \mod p(x)$ and then $(x^2 + 1)y \cong 1 \mod x^4 + 2$. That gets me nowhere. However $yx^2 + y - 1 \cong x^4 - 2$. Do I have to find linear factors now?

• On the first one, if you are moding out by p(x) then p(x) factors as 0 in F. In fact, p(x) has a root in $Z_3$. – Vladhagen Nov 26 '13 at 14:08
• Can you go into more detail @Vlad? I understand $p(1) = 0$. Not $p(2)$ or $p(3)$ though. – Don Larynx Nov 26 '13 at 14:09
• As written the factorization is p(x)=0. If you are looking at all polynomials mod p(x), then p(x) itself is zero. – Vladhagen Nov 26 '13 at 14:10
• It factors as zero. That is how modulo arithmetic works in a polynomial ring. A much better question would ask how the polynomial factors in $Z_3[x]$. Then $p(x) = (x+2)(x^2+x+2)$ – Vladhagen Nov 26 '13 at 15:14
• When you mod by p, you get p is zero. It is sort of like asking how 24 factors mod 24. It just factors as zero. – Vladhagen Nov 26 '13 at 15:16