Vector space problem Do these sets make a vector space :
1) $$M_1 = \bigl\{(x,y)\in\mathbb{Q}^2\mid 80996x + 40498y = 0\bigr\}$$
2) $$M_2 = \bigl\{F(x)\in\mathbb{R}[x]\mid\deg\bigl(F(x)\bigr) < 4; F(80996) = 0\bigr\}$$
Can you show me how to do these sort of exercises and if someone could also direct me to some literature and exercises of this kind I would be forever great full. I really want to learn how to do these problems on my self, but unfortunately I can't find the right stuff on the web.Again thanks to anyone who can help.
 A: Since $\mathbb Q^2$ and $\mathbb R[x]$ are both vector spaces, all that's left to answer is whether the sets you've defined are closed under addition and scalar multiplication.  Any subset of a vector space that is closed under addition and scalar multiplication is itself a vector space.  Another way of expressing "closed under addition and scalar multiplication" is "closed under linear combinations".  So is every sum of two members of $\bigl\{F(x)\in\mathbb{R}[x]\mid\deg\bigl(F(x)\bigr) < 4; F(80996) = 0\bigr\}$ itself a member of that set?  Call those two members $f(x)$ and $g(x)$.  Both are polynomials of degree $<4$.  Is their sum a polynomial of degree $<4$?  Is every scalar multiple of $f(x)$ a polynomial of degree $<4$?  If $f(80996)=0$ and $g(80996)=0$, then is $(f+g)(80996)=0$?  If $f(80996)=0$, then is every scalar multiple of $f(x)$ evaluated at $80996$ equal to $0$?
For $M_1$, if the pair $(x_1,y_1)\in\mathbb Q^2$ satisfies $80996x + 40498y = 0$ and so does the pair $(x_2,y_2)$, then does the pair $(x_1,y_1)+(x_2,y_2)$ satisfy that equation?  In this case, I'm guessing "scalar" would mean "rational number", so does $r(x_1,y_1)$ satisfy that equation whenever $r$ is rational?
