What are the elements like in $\mathbb C[x]/(2x^2+5)$?

Question

As stated, I wonder what are the elements in $\mathbb{C}[x]/(2x^2+5)$ composed like? Since $2x^2+5$ is not monic, it seems to be a little different from the situation like in $\mathbb C[x]/(x^2+5/2)$. Please help me.

Question Complement: Thank you, but maybe I chose a not that good polynomial ring: $\mathbb C[x]$, so it seems some kind of trivial. What if I change it by $\mathbb Z[x]$, $\mathbb Q[x]$, or other polynomial rings?

Answer 1

Since $\mathbb C$ is algebraically closed, the polynomial $x^2+5/2$ splits as $(x-\sqrt{5/2}i)(x+\sqrt{5/2}i)$. Thus the quotient ring $\mathbb C[x]/(x^2+5/2)$ is not an integral domain, but splits as a direct product (because the ideals are coprime). Indeed, $\mathbb C[x]/(x^2+5/2) \approx \mathbb C\times \mathbb C$.

To get a hint of why this is true, noe that $(x-\sqrt{5/2}i) \cdot (x+\sqrt{5/2}i) \equiv 0$ in the quotient, just as $(1,0)(0,1)=0$ in $\mathbb C \times \mathbb C$.

Edit: For other polynomial rings, the following is true: if you replace $\mathbb C$ by a field $K$, then the answer depends on whether or not $2x^2+5$ factors or not. This happens if and only if $\sqrt{5/2}i$ is an element of $K$. (be ware of the cases $\mathrm{char } K=2,5$. If $\sqrt{5/2}i \not \in K$, then the quotient is isomorphic to $K[\sqrt{5/2}i] \approx K \oplus \sqrt{5/2}i.K$

Notice that in both cases, the quotient is a vector space of dimension over $K$ equal to the degree of the polynomial. It is just whether or not it has zero-divisors that depend upon the splitting of the polynomial.

Answer 2

Note that $I = \langle 2x^2 + 5 \rangle = \langle x^2 + \frac{5}{2} \rangle$ because if $$ g(x) \in \langle x^2 + \frac{5}{2} \rangle \implies g(x) = h(x) \left(x^2 + \frac{5}{2} \right) = \frac{1}{2}h(x) ( 2x^2 + 5 ) \implies g(x) \in \langle 2x^2 + 5 \rangle $$ and you can do the opposite inclusion similarly. Note that this only works because $\frac{1}{2} h(x) \in \mathbb{C}[x]$ for any $h(x) \in \mathbb{C}[x]$.

From here we can conclude that $$ \mathbb{C}[x] / \langle 2x^2 + 5 \rangle = \mathbb{C}[x] / \langle x^2 + \frac{5}{2} \rangle $$ and you can work from here to understand the quotient.

EDIT: Note that this is not a field since $x^2 + \frac{5}{2}$ is not irreducible in $\mathbb{C}[x]$.