Verification that a vector space (specified below) is a direct sum of two subspaces (This is a question from Axler's "Linear Algebra Done...

Verification that a vector space (specified below) is a direct sum of two
subspaces (This is a question from Axler's "Linear Algebra Done...

I'm working through Axler's "Linear Algebra Done Right". On page 15, it
gives this example:
Consider the vector space $P(F)$ of all polynomials with coefficients in
$F$ (ie., coefficients taken from the real numbers or complex numbers).
Let $U_e$ denote the subspace of $P(F)$ consisting of all polynomials $p$
of the form: $$p(z) = a_0 + (a_2)z^2 + ... (a_2m)z^{2m}$$ and let $U_0$
denote the subspace of all polynomials $p$ of the form: $$p(z) = (a_1)z +
(a_3)z^3 + ... + (a_{2m+1})z^{2m+1};$$ $m$ is a nonnegative integer and
the coefficients are from the reals (to keep things simple).
Next, it says I should verify that $P(F)$ is a direct sum of $U_e \oplus
U_0$.
Well, to do this I wanted to use a theorem that occurs a few pages later,
namely:
if $U_1 ... U_n$ are subspaces of $V$. Then $V$ is a direct sum of $U_1
... U_n$ iff:
a. $V = U_1 + ... U_n$
b. the only way to write $0$ as a sum $u_1 + ... + u_n$, where each $u_j$
is in $U_j$, is by taking all the $u_j$ in $U_j$, is by taking all the
$u_j$'s equal to $0$.
I wanted to use this theorem to show $P(F)$ is a direct sum of $U_e \oplus
U_0$. (a) -- from the above theorem -- is clearly satisfied. What has me
stumped in part (b). It seems to me that that are multiple ways to write
the $0$ for the sum of $u_e + u_0$, $u_e \in U_e$, $u_0 \in U_0$. One way
to get $0$ is make each coefficient $0$. Or you could make sure that each
entry from $U_e$ is paired off with the corresponding entry from $U_0$,
and that together they add up to $0$. When $U_e$ and $U_0$ are added
together, this would give a $0$ too.
But this can't be right b/c then $P(F)$ would not be a direct sum of $U_e$
and $U_0$.
Any help?