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Let $I = _{\mathbb {C} [X]} \langle X^2 + 1\rangle$ the principal ideal of $\mathbb{C}[X]$ generated by $X^2 + 1$. Is $\mathbb{C}[X]/I$ an integral domain?

From my understanding $\mathbb{C}[X]/I$ is the set of all cosets of $\langle X^2 + 1\rangle$ in $\mathbb{C}$. Now to me, since $X^2+ 1$ generates all of $\mathbb{C}$, and since $\mathbb{C}$ is an integral domain so is $\mathbb{C}[X]/I$. However, intuition is telling me I have misunderstood $\mathbb{C}[X]/I$ can someone help please?

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Factor $x^2 + 1$ over $\mathbb{C}$ to suggest two zero divisors in the quotient ring. – user43208 Oct 18 '13 at 20:43
I am wanting to say $i$ and $-i$ are zero divisors. $(i^2 + 1)(b) = 0$ for any $b$, $b(i^2 + 1) = 0$ for any $b$ ? (similarly for $-i$) – yhu Oct 18 '13 at 20:56
No, those are not the zero divisors (nor are they zero divisors). I meant my hint even more literally than that. What are the polynomial factors? – user43208 Oct 18 '13 at 20:59
$(x + i)(x - i)$ – yhu Oct 18 '13 at 21:02
Yes, and now consider those factors modulo the ideal $(x^2+1)$, which represents zero in the quotient ring. – user43208 Oct 18 '13 at 21:03

From the comments below the OP, I think we should get back to fundamentals.

The elements of a quotient ring $R/I$ may be taken to be cosets $a + I$ of the ideal $I$. Two cosets $a + I$, $b + I$ are equal iff $a - b \in I$. Cosets are added and multiplied according to the rules $(a + I) + (b + I) := (a + b) + I$ and $(a + I)(b+I) := ab + I$. The addition and multiplication thus defined provide the set of cosets $a + I$ with a ring structure, whose zero element is $0 + I = I$. This describes the quotient ring $R/I$.

In the present case, $I = (x^2 + 1)$ (the ideal $I$ is $\{(x^2 + 1)p(x): p(x) \in \mathbb{C}[x]\}$). Again, this ideal is the coset $0 + I$ that represents the zero element of the quotient ring.

We have $(x+i)(x-i) = x^2 + 1$. The hint was trying to suggest that this factorization provides zero divisors, in the quotient ring. What that means is that the factors $x+i$ and $x-i$ should suggest what cosets to try, whose product will be zero in the quotient ring.

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Thank you so much. I was so muddled with all of the definitions, and you have provided them in a logical manner. Hopefully this will mean I can solve future problems without 101 questions! – yhu Oct 18 '13 at 21:37
You're very welcome. Please consider upvoting and/or accepting this as an answer, if it indeed answered your question. – user43208 Oct 18 '13 at 21:39

It's probably easiest to think of ${\mathbb C}[X]/(X^2 + 1)$ as the set of all expressions of the form $a x + b$ with $a, b \in {\mathbb C}$ with the special property that $x^2 = -1$.

Then, working in ${\mathbb C}[X]/(X^2 + 1) = {\mathbb C[x]}$ with $x^2 = -1$, $(x - i)(x + i) = x^2 - i^2 = x^2 + 1 = 0$, so $x - i$ and $x + i$ are zero-divisors.

Or, avoiding going to the quotient ring at all, $X^2 + 1 = (X + i)(X - i)$, so the ideal generated by $X^2 + 1$ is not a prime ideal and therefore ${\mathbb C}[X](X^2 + 1)$ is not an integral domain.

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