2
$\begingroup$

Let $R=\mathbb{R}[x]$ be the ring of polynomials and $I=(x^2+1)$, $J=(x+1)$ and $K=(x^3+x^2+x+1)$ ideals of $R$.

It's easy to see that there is a ring isomorphism between $R/K$ and $R/I \times R/J$. Find the element in $R/K$ whose image in $I \times J$ is $(2x+1+I,3+J)$.

How do I find this element? I know how to apply the Chinese remainder theorem in groups with $\mathbb{Z}/n\mathbb{Z}$, but I don't understand how I'm supposed to find this element in this case.

$\endgroup$
1
$\begingroup$

Let $p(x)$ and $q(x)$ be elements of $R$ whose associated principal ideals are comaximal. Then there exist polynomials $a(x)$ and $b(x)$ in $R$ such that $ a(x)p(x) + b(x)q(x) = 1$ and we can use a process based on Euclid's algorithm for greatest common divisor (repeated division with remainder) to find these polynomials.

Certainly $a(x)p(x)\equiv 0$ mod $(p(x))$. More interesting, $a(x)p(x)\equiv 1$ mod $(q(x))$. Similarly, $b(x)q(x)\equiv 1$ mod $(p(x))$ and $b(x)q(x)\equiv 0$ mod $(q(x))$.

For this problem, take $p(x)= x^2+1$ and $q(x)=x+1$. The Euclidean algorithm requires only one division with remainder to give $x^2+1 = (x+1)(x-1) + 2$ from which we conclude that $$ \left(\frac{1}{2}\right) (x^2 + 1) - \left(\frac{x-1}{2} \right)(x+1) = 1$$

The polynomial $$ f(x) = -(2x+1) \left(\frac{x-1}{2} \right)(x+1) + 3 \left(\frac{1}{2}\right) (x^2 + 1)$$

is congruent to $(2x+1)$ mod $I$ and congruent to $3$ mod $J$. So $f(x)+K$ is the inverse image of $(2x+1+I, 3+J)$ that you seek.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.