Prove that $\mathbb{C}[x]/(x^2+3x)$ is isomorphic to $\mathbb{C} \oplus \mathbb{C}$, where $\mathbb{C} \oplus \mathbb{C} = \{(z_1, \space z_2) \mid z_1, z_2 \in \mathbb{C} \}$ is the ring with componentwise operations of addition and multiplication.
Can you help me with understanding of what $\mathbb{C} \oplus \mathbb{C}$ actually is and what stands for $\mathbb{C}[x]/(x^2+3x)$, this task was given to me as training one before final test, however our course was fairly compressed due to pandemic, so would appreciate it if anyone would explain how to solve this task or even write down solution for me to explore it fully.
P.S: Have been thinking of
$ \varphi\colon\mathbb{C}[x]\to\mathbb{C}\oplus\mathbb{C}
\qquad
\varphi(f)=(f(0),f(-3))
$, however have no idea of how to prove this and even have I understood task properly.
Best Answer
The ring $A \oplus B$ is the direct product of the rings $A$ and $B$. This means the elements are pairs $(a, b)$ with $a \in A$ and $b \in B$. Ring operations are done pointwise, i.e., $(a, b) + (c, d) = (a + c, b + d)$ and $(a, b)\cdot(c, d) = (a\cdot c, b \cdot d)$. So $\mathbb C \oplus \mathbb C$ is just pairs of complex numbers with addition and multiplication done pointwise.
The ring $\mathbb C[x]/(x^2 + 3x)$ is the quotient of the ring $\mathbb C[x]$ by the principal ideal $(x^2 + 3x)$. For ease of notation let $I = (x^2 + 3x)$. The elements are additive cosets $f(x) + I$ (the polynomial $f$ is called a representative). You test for equality of cosets by testing if the difference of representatives is in the ideal, so $f(x) + I = g(x) + I$ if and only if $f - g \in I$. Finally the ring operations are done on the representative, so $(f + I) + (g + I) = (f + g) + I$ and $(f + I)(g + I) = (fg) + I$. Another way of describing this ring is as a ring of remainders. It's elements are the possible remainders one can get when dividing a polynomial by $x^2 + 3x$, so the elements are all degree $0$ or degree $1$ polynomials. The ring operations are the usual operations from $\mathbb C[x]$ but with a final step of dividing the result by $x^2 + 3x$ and taking the remainder. Hopefully your class covered both of these descriptions and proved that they are equal.
The map you suggest, $f \mapsto (f(0), f(-3))$, in your p.s. is indeed the correct map so if the structural approach using the chinese remainder theorem doesn't appeal to you then you could alternatively just prove that the map you've written down is a ring homomorphism and is surjective and injective.
For surjective, if you prove that the map is a ring homomorphism then in particular it is linear as a map between vector spaces over $\mathbb C$. Then for the image to be all of $\mathbb C \oplus \mathbb C$ you just need to show that there are two linearly independent vectors in the image. If you just choose two random polynomials and map them over then I think it's pretty likely you'll get independent images just at random so that shouldn't be too hard. Alternatively, you could design polynomials that you know will have linearly independent images by making sure that the images are zero in one coordinate and nonzero in the other.
For injective use the fact that $f(a) = 0$ if and only if $x - a$ divides $f$.