Let $f=x^2+x+\overline{9} \in \mathbb Z_{11}[x]$. Show that $I=\left\{sf\mid s\in \mathbb Z_{11}[x] \right\}$ matches $J=\left\{h \in \mathbb Z_{11}[x] \mid h(\overline{1}) = h(\overline{-2}) = \overline{0}\right\}$. I don't have the palest idea where to start, can you please push me in the right direction?
Asked
Active
Viewed 61 times
2
-
Are $I$ and $J$ supposed to be ideals of $\mathbb{Z}_{11}[x]$? If so, something is funny. – Mar 10 '13 at 14:58
-
1In the definition of $I$, don't you mean $s\in\mathbb{Z}_{11}[x]$? – akkkk Mar 10 '13 at 15:19
-
Sorry guys I messed up while copying. That $h$ is indeed an element in $\mathbb Z_{11}[x]$. – haunted85 Mar 10 '13 at 15:19
-
$f(\overline 2)=\overline 4$ so $f\not\in J$ but $f\in I$ with $s=1$. So $I\neq J$ – akkkk Mar 10 '13 at 15:23
3 Answers
1
Note that $F=\mathbb{Z}_{11}$ is a field (although actually, the only thing you need is that $-3=8$ is invertible.
Then recall that for a polynomial $f\in F[X]$ and $a\in F$ $$ f(a)=0\quad\Leftrightarrow \quad f(x)=(x-a)g(x) $$ for some $g\in F[X]$.
Apply this to $f\in J$ and $a=1$. This gives $$ f(x)=(x-1)g(x). $$ Now $f(-2)=0=-3g(-2)$ so $g(-2)=0$ since $-2-1=-3\neq 0$ is invertible in $F$. So $$ g(x)=(x+2)h(x). $$ Finally $$ f(x)=(x-1)(x+2)h(x)=(x^2+x-2)h(x)=(x^2+x+9)h(x). $$ So $J$ is contained in $I$. And the converse is trivial.
Julien
- 44,791
0
Can you see how every element $x\in I$ has $x(\overline 1)=x(-\overline 2)=0$? If $\overline 1$ and $-\overline 2=\overline 9$ are zeros of $h\in J$, can you give a factor of $h$?
akkkk
- 1,889
-
It is not obvious to me, please can you explain how you are able to tell that every element $x \in I$ has $\overline{1}$ and $\overline{-2}$ as roots? – haunted85 Mar 10 '13 at 15:39
-
0
Hint $\ $ Notice $\rm\ mod\ 11\!:\ x^2\!+x+9 \,\equiv\, (x-1)(x+2),\ $ then apply the following
Bifactor Theorem $\ $ Let $\rm\,a,b\in R,\,$ a ring, and $\rm\:f\in R[x].\:$ If $\rm\ \color{#C00}{a\!-\!b}\ $ is cancelable in $\rm\,R\,$ then
$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in R[x]$$
Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying Factor Theorem twice, canceling $\rm\: \color{#C00}{a\!-\!b},\:$ we obtain
$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in R[x]\\ \rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in R[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$
Remark $\ $ The theorem may fail when $\rm\ a\!-\!b\ $ is not cancelable (i.e. is a zero-divisor), e.g.
$\rm\quad mod\ 8\!:\,\ f(x)=x^2\!-1\,\Rightarrow\,f(3)\equiv 0\equiv f(1)\ \ but\ \ x^2\!-1\not\equiv (x\!-\!3)(x\!-\!1)\equiv x^2\!-4x+3$
Math Gems
- 19,574