Computing the inverse of an element in $\mathbb{Z}_5[x] / \langle x^2+x+2\rangle$

Question

How does one calculate the inverse of $(2x+3)+I$ in $\mathbb{Z}_5[x] / \langle x^2+x+2\rangle$?

Give me some hint to solve this problem.

Thanks in advance.

Answer 1

The generator $x^2 + x + 2$ of the ideal is quadratic, so every element admits an (in fact unique) linear representative, $$ ax + b . $$ We want to find the element $ax + b + I$ such that $$[(2x + 3) + I][(ax + b) + I] = 1 + I.$$

Simplifying gives $$ 2a x^2 + (3a + 2b) x + (3b - 1) \in I.$$

Now, we can eliminate the quadratic term by subtracting from the left-hand side the element $2a(x^2 + x + 2) \in I$ without changing the equation, and hence produce the unique linear representative in its equivalnce class: $$(a + 2b) x + (-4a + 3b + 1) \in I.$$ Since this element is linear and in $I$, it is zero: $$(a + 2b) x + (-4a + 3b + 1) = 0.$$ This reduces the problem to solving an (inhomogeneous) $2 \times 2$ linear system over $\mathbb{Z}_5$.

Answer 2

You need $$(2x+3)p(x)\equiv1\pmod{x^2+x+2}$$ in ${\Bbb Z}_5[x]$. That is, $$(2x+3)p(x)+(x^2+x+2)q(x)=1\ ,$$ and you can solve this by using the Euclidean algorithm (not forgetting that all the arithmetic of coefficients is modulo $5$).

If you have seen how to use the Euclidean algorithm in $\Bbb Z$ to solve equations such as $23x+112y=1$, this problem is very similar.