Quotient Ring?

Algebra Level 5

Let $\mathbb R[X]$ be the polynomial ring of real coefficients. Then there exists some ideal $J$ such that the quotient ring $\mathbb R[X]/J\cong$ :

$(\mathbb R,+,\times)$ , the real numbers ring
$(\mathbb C,+,\times)$ , the complex numbers ring
$(\mathbb H,+,\times)$ , the quaternion ring

1 only 2 only 3 only 1, 2 only 1, 3 only 2, 3 only All of them None of them

1 solution

展豪張
Jun 22, 2016

Let $\langle X\rangle$ be the ideal generated by $X$
Then $\mathbb R[X]/\langle X\rangle\cong(\mathbb R,+,\times)$ , since the ideal $\langle X\rangle$ 'wiped out' all the terms with $X$ , leaving the 'constant term'.
Let $\langle X^2+1\rangle$ be the ideal generated by $X^2+1$
Then $\mathbb R[X]/\langle X^2+1\rangle\cong(\mathbb C,+,\times)$ , since the ideal makes $X^2+1\sim0$ , which makes $X$ performs like the imaginary unit $i$ .
Polynomial ring is an abelian structure while quaternion is not. Quaternion can never be generated in this way. (Though it can be formed from free algebra, which does not assume commutativity.)

Nice solution.

Hana Wehbi - 4 years, 11 months ago

Thank you! =D

展豪張 - 4 years, 11 months ago

$\mathbb{R}$ is a field, if and only if, $\mathbb{R}[X]$ is an euclidean domain and this implies that $\mathbb{R}[X]$ is a PID.

1.- Let $g:\mathbb{R}[X] \to \mathbb{R}$ such that $g(a_nx^n + a_{n -1}x^{n - 1} + ... + a_1x + a_0) = a_0$ then g is an epimorphism of rings and this implies $\frac{\mathbb{R}[X]}{\text{Ker g}} \cong \mathbb{R}.$ Since $\mathbb{R}[X]$ is a PID it is easy to see $\text{Ker g} = \langle X \rangle$ and applying the first theorem of isomorphy we get $\mathbb{R}[X] / \langle X \rangle \cong (\mathbb{R}, +, \times)$ .

2.- $\frac{\mathbb{R}[X]}{\langle X^2 + 1 \rangle} = \{ aX + b + \langle X^2 + 1\rangle; \space a, b \in \mathbb{R}\} \cong (\mathbb{C}, +, \times)...$

3.- ...

P.S.- It was for developing it a little bit :)... Folk, don't push the downvote key?haha...

Guillermo Templado - 4 years, 11 months ago

Nice solution, specially 3rd :-)

akash patalwanshi - 3 years, 3 months ago

Quotient Ring?

1 solution

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