Show that the map is a linear functional and determine the dimension of the quotient space $V/ker(y)$

Question

During an exam I was given the following task:
Consider the vector space $$V:=\{p \in \mathbb{Q}[x]\ |\ p \text { has degree at most } 3\}$$ Show that the map $$y(p)=p(0),\quad V\rightarrow\mathbb{Q}$$

is a linear functional and determine the dimension of the quotient space $V/ker(y)$. I argued that: $$y\left(\gamma_{1} p_{1}+\gamma_{2} p_{2}\right)=p(0)=\sum_{i=0}^{n} \alpha_{i} 0^{i}=0$$ and $$\gamma_{1} y\left(p_{1}\right)+\gamma_{2} y\left(p_{2}\right)=\gamma_{1} p(0)+\gamma_{2} p(0)=\gamma_{1} \sum_{i}^{n} \alpha_{i} 0^{i}+\gamma_{2} \sum_{i}^{n} \alpha_{i} 0^{i}=0$$ However, I am not sure if this is correct, because what is $0^0$?
To determine the dimension of $V/ker(y)$, I argued that $y$ is the zero map with dimension $0$ and that the dimension of a polynomial of degree at most $3$ is $4$, and therefore that the dimension of $V/ker(y)$ is $$\operatorname{dim} V-\operatorname{dim}(\operatorname{ker} y)=4-0=4$$

Will someone tell me if I messed up?

Answer 1

In such expressions, $0^0$ is defined to be $1$.

Clearly $y$ is NOT the zero map. For example, the polynomial $1 \in V$ gets mapped to $1$. Also $X+1$ gets mapped to $1$.

By the isomorphism theorem, we have $$V/\ker y \cong y(V)= \mathbb{Q}$$

and thus $\dim_\mathbb{Q}(V/\ker y) = 1$.

Answer 2

No, it is not correct.

You are supposed to check that, if $\gamma_1,\gamma_2\in\Bbb Q$ and $p_1,p_2\in V$, then $y(\gamma_1p_1+\gamma_2p_2)=\gamma_1y(p_1)+\gamma_2y(p_2)$. However, you wrote that $y(\gamma_1p_1+\gamma_2p_2)=p(0)$, without saying what $p$ is. If$$p_1(x)=\alpha_0+\alpha_1x+\alpha_2x^2+\alpha_3x^3\text{ and }p_2(x)=\beta_0+\beta_1x+\beta_2x^2+\beta_3x^3,$$then both $y(\gamma_1p_1+\gamma_2p_2)$ and $\gamma_1y(p_1)+\gamma_2y(p_2)$ are equal to $\alpha_0+\beta_0$. That's all.

Besides, $\dim\ker y=3$, not $4$.