I'm given a polynomial ring $S=K[x_1,...,x_n]$ and $I$ is an ideal of $S$.
I'm working on proving that the quotient ring $S/I$ is a vector spake over $K$. Since S is a ring, we already have some of the vector space properties, such as closure under addition, multiplication and distributivity, etc. I need to show two additional properties for $S/I$ to be a vector space.
The first property is $\forall u,v\in S/I,(u+v)\in S/I$
My attempt at this is by letting $u=a+I$ and $v=b+I$, where $a,b\in S$. Then $u+v=(a+I)+(b+I)$, by the associative property on rings, $u+v=(a+b)+(I+I)$. by closure under addition, $a+b\in S$, but then I need to show that $I+I=I$. Can I say that $I+I=\{x+y|x,y\in I\}$, which is clearly equal to $I$, since $0\in I$, so setting either $x$ or $y$ to $0$, we can find every element in $I$ inside $I+I$. But isn't $I+I=2I=\{2x|x\in I\}\neq I$.
For the second property: $\forall u\in S/I$ and $c\in K$, $cu\in S/I$
Similarly, I let $u=a+I$ for some $a\in S$. So $cu=c(a+I)=ca+cI$ by the distributive law of fields. $ca\in S$, since $S$ is a ring. But then how do I show that $cI=I$.
This is the curse of too much information.
Theorem If $F$ is a field and $\varphi : F \to R$ is a ring homomorphism, then $R$ is a vector space over $F$ with scalar multiplication $f \cdot r = \varphi(f) r$.
Your problem follows, using the composite $K \to S \to S/I$.
(the converse is true too: if a ring $R$ has the structure of an $F$-vector space that uses the same addition operation, then there is a ring homomorphism $F \to R$ given by $\varphi(f) = f \cdot 1_R$)