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Find the multiplicative inverse of $x + 2$ in the field $\Bbb Z_5[x]/(x^2 + 2)$.

I have done the following so far:

\begin{align*} x^2+2 &= (x+2)(x+3) + 1\\ (x+2)(x+3) &\equiv -1 \pmod {x^2+2}\\ (x+2)(x+3) &\equiv 4 \pmod {x^2+2}, \end{align*}

therefore multiplying both sides by $4,$

$$4(x+2)(x+3) \equiv 1 \pmod {x^2+2}.$$

This is as far as i can get to figuring it out.

Now I have the final answer and it states that: inverse of $(x+2) = 4(x+3) = 4x+2$??? Can someone please explain to me how the answers are able to go from where i have finished to this final solution?

Thank you

pjs36
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Oliver
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  • You were finished. By your calculation, $[x+2][4(x+3)]\equiv 1\pmod{x^2+2}$. So $4x+2$ is the inverse of $x+2$. – André Nicolas Jun 13 '15 at 23:53
  • I think it would be more expressive to write the elements of $\mathbf F_5$ as $0,\pm 1,\pm2$. – Bernard Jun 13 '15 at 23:59
  • @AndréNicolas, can you please explain it in a little more detail as to how 4x+2 is the inverse, considering initially i have [x+2][4(x+3)]. How is the 4 not related to (x+2) initially? – Oliver Jun 14 '15 at 00:08
  • I do not know what the issue is. In arithmetic mod $5$, we have $4(x+3)=4x+2$ since $4\cdot 3=2$. Or is the uncertainty somewhere else? – André Nicolas Jun 14 '15 at 00:13
  • Sorry there is some uncertainty elsewhere, i may not have not explained myself properly. When i multiply both sides by 4 to result in 4((x+2)(x+3)) how does that lead to what your responded with of only multiply (x+3) by 4. [x+2][4(x+3)]? – Oliver Jun 14 '15 at 00:19
  • Either use the extended Euclidean algorithm as in the linked dupe, or rationalize the denominator as below, where we use $\sqrt{-2}$ to denote the image of $,x,$ in $,\Bbb Z_5[x]/(x^2!+2)\cong \Bbb Z_5[\sqrt{-2}].,$ See also comments here

    $$\dfrac{1}{2+\sqrt{-2}} ,=, \dfrac{1}{2+\sqrt{-2}},\dfrac{2-\sqrt{-2}}{2-\sqrt{-2}} ,=, \dfrac{2-\sqrt{-2}}{6} ,=, 2-\sqrt{-2}\qquad\qquad$$

     – Bill Dubuque May 27 '24 at 20:49
  • $(x+2)f(x)\equiv 1\Rightarrow f(x)\equiv (x+2)^{-1}$ by definition / uniqueness of inverses. $\ \ $ – Bill Dubuque May 27 '24 at 20:54

2 Answers2

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Another way to compute the inverse of $x+2$ in $K={\mathbb{Z}_5[x]/(x^2+2)}$ is to use the fact that every element in $K$ is represented by a polynomial in $x$ of degree $\leq1$, i.e. $$ \frac1{(x+2)}=ax+b $$ for some $a$, $b\in\mathbb{Z}_5$. Thus we can think of the identity $$ (x+2)(ax+b)=1 $$ as a system of equations over $\mathbb{Z}_5$. Namely (using the fact that $x^2=-2=3$ in $K$) $$ 2a+b=0\qquad 2(b-a)=1 $$ which after some straightforward manipulation gives $a=4$ and $b=2$.

Andrea Mori
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$b$ is said to be inverse of $a$ if $ab\equiv1\pmod n$. Here you find out $4(x+2)(x+3) \equiv 1 \pmod {x^2+2}.$ It means $4(x+3)$ is the inverse of $(x+2)$ and $4(x+3)=4x+12$ which is equal to $4x+2$ under modulo $5$.

Maths Wizard
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