Can we say that the ring $R = \frac { \mathbb C[x] }{<x^2+1>}$ is a Principal Ideal ring?

Question

Can we say that the ring $R = \frac{\mathbb C[x] }{<x^2+1>}$ is a Principal Ideal ring?

If we can say that elements except $<x^2+1>$ in every ideal in $ \mathbb C[x]$ which contains $x^2 +1$ is also an ideal in $R = \frac{\mathbb C[x] }{<x^2+1>}$ then we are able to say.

But can we say the last line.

If it is not true , can anyone please point out the error ? I am a first reader of Algebra. Please forgive me if the question is too easy to be posted.

Answer 1

By the Chinese remainder theorem, using $x^2+1=(x-i)(x+i)$, we get that $$ R=\Bbb C[x]/(x^2+1)\\ \cong\Bbb C[x]/(x-i)\times \Bbb C[x]/(x+i)\\ \cong \Bbb C^2 $$ This ring only has four ideals, and each of them is easily checked to be principal.

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More details:

The Chinese remainder theorem. This is a rather large topic all to itself, but here are some details on what happens in this specific case.

Since $x - i + (x^2 + 1)$ is an element of $\Bbb C[x]/(x^2 + 1)$, there is a quotient map $$ \Bbb C[x]/(x^2 + 1)\to\frac{\Bbb C[x]/(x^2 + 1)}{(x - i + (x^2 + 1)}\cong \Bbb C[x]/(x-i) $$ Similarily, there is a quotient map $$ \Bbb C[x]/(x^2 + 1)\to\frac{\Bbb C[x]/(x^2 + 1)}{(x + i + (x^2 + 1)}\cong \Bbb C[x]/(x +i) $$ The isomorphisms here are given by the third isomorphism theorem: Given a ring $R$ with an ideal $I$, an ideal $J\subseteq R/I$ with corresponding ideal $J'\subseteq R$, then $(R/I)/J\cong R/J'$.

These two homomorphisms together give a homomorphism $$ \Bbb C[x]/(x^2 + 1) \to \Big(\Bbb C[x]/(x-i), \Bbb C[x]/(x+i)\Big)\cong \Bbb C^2 $$ It turns out that this homomorphism is an isomorphism.

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The four ideals of $\Bbb C^2$ are $$ \{0\}\times \{0\} = ((0,0))\\\Bbb C\times \{0\} = ((1, 0))\\ \{0\}\times \Bbb C = ((0,1))\\ \Bbb C\times \Bbb C = ((1,1)) $$ It's not difficult to show that any generating set generates one of these four ideals.

Answer 2

It is true. If $R$ is a PID and $I$ is an ideal in $R$, then the quotient ring $R/I$ is a principal ideal ring as well.

Indeed, take an ideal $K$ of $R/I$. By the correspondence theorem $K = H/I$ where $H$ is an ideal of $R$ that contains $I$. But $H$ is principal. I.e. $H = (r)$ for some $r \in R$, and it follows that $K = (r)/I = (r+I)$ is principal (here I denote an element of the quotient ring with representant $r$ as $r+I$).