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Recently, I asked a question about axoims of vector spaces and counterexamples. The question can be found at: A Counter Examples in Linear Algebra (Vector Space)

However, a few days before asking this question, I tried to prove that the axiom $1 cdot v = v$ in the definition of a vector space is redundant. For the proof, I have used that every vector space has a basis (which is proved by using Zorn's Lemma). However, I know that this is not true. Because we do have a system where every other axiom is satisfied and only $1 cdot v neq v$ and hence the system fails to be a vector space. I would like to know where exactly in the proof have I made a mistake? I have been stuck on this for a few weeks and no matter what I do, I cannot find the exact point where things start to go wrong. All help, suggestions and comments are appreciated. The proof I tried is as follows:-

We know that $V$ must have a basis (can be proved by using Zorn's Lemma), say $B$. Let $v in V$. Then, $exists v_1, v_2, cdots, v_n in B$ and $alpha_1, alpha_2, cdots, alpha_n in mathbbF$ such that $v = alpha_1 cdot v_1 + alpha_2 cdot v_2 + cdots + alpha_n cdot v_n = sumlimits_i = 1^n alpha_i cdot v_i$. Now, let $1 cdot v = w$, where $1 in mathbbF$ is the unity of the field and $w in V$. Again, since $B$ is a basis, $exists w_1, w_2, cdots, w_m in B$ and $beta_1, beta_2, cdots, beta_m in mathbbF$ such that $w = beta_1 cdot w_1 + beta_2 cdot w_2 + cdots + beta_m cdot w_m = sumlimits_i = 1^m beta_i cdot w_i$. Therefore, we have
beginalign*
1 cdot sumlimits_i = 1^n alpha_i cdot v_i &= sumlimits_i = 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^n 1 cdot left( alpha_i cdot v_i right) &= sumlimits_i = 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^n left( 1 alpha_i right) cdot v_i &= sumlimits_i = 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^n alpha_i cdot v_i &= sumlimits_i = 1^m beta_i cdot w_i \
endalign*
Consider the two sets $S_1 = leftlbrace v_1, v_2, cdots, v_n rightrbrace$ and $S_2 = leftlbrace w_1, w_2, cdots, w_m rightrbrace$. Clearly, $S_1 subseteq B$ and $S_2 subseteq B$ and hence $S_1, S_2$ are linearly independent. Also, the set $S_1 cup S_2 = leftlbrace v_1, v_2, cdots, v_n, w_1, w_2, cdots, w_m rightrbrace subseteq B$ and is also linearly independent. Let, if possible $S_1 cap S_2 = emptyset$, where $emptyset$ denotes the empty set. This means that $forall i in leftlbrace 1, 2, cdots n rightrbrace$ and $forall j in leftlbrace 1, 2, cdots, m rightrbrace, v_i neq w_j$.

Using that the additive inverse of $a cdot v$ is $left( -a right) cdot v$ and adding the additive inverses of each vector $beta_i cdot w_i$ in the last equation to obtain,
$$sumlimits_i = 1^n alpha_i cdot v_i + sumlimits_i = 1^m left( - beta_i right) cdot w_i = textbf0$$
Since $S_1 cup S_2$ is linearly independent, $forall i in leftlbrace 1, 2, cdots, n rightrbrace, alpha_i = 0$ and $forall i in leftlbrace 1, 2, cdots, m rightrbrace, - beta_i = 0$, which in turn gives that $beta_i = 0$. Therefore, $v = 0 cdot v_1 + 0 cdot v_2 + cdots + 0 cdot v_n = textbf0$ and $w = 0 cdot w_1 + 0 cdot w_2 + cdots + 0 cdot w_m = textbf0$. Hence, $1 cdot textbf0 = textbf0$.

Now, let us consider that $S_1 cap S_2 neq emptyset$. Let there be $r$ vectors, where $0 leq r leq min leftlbrace m, n rightrbrace$, which are common in $S_1$ and $S_2$. We shall name them, $v_i$, where $i in leftlbrace 1, 2, cdots, r rightrbrace$. Thus, our sets look like $S_1 = leftlbrace v_1, v_2, cdots, v_r, v_r + 1, cdots, v_n rightrbrace$ and $S_2 = leftlbrace v_1, v_2, cdots, v_r, w_r + 1, cdots, w_m rightrbrace$. Now, $v = sumlimits_i = 1^r alpha_i cdot v_i + sumlimits_i = r + 1^n alpha_i cdot v_i$ and $w = sumlimits_i = 1^r beta_i cdot v_i + sumlimits_i = r + 1^m beta_i cdot w_i$. Again,
beginalign*
1 cdot left( sumlimits_i = 1^r alpha_i cdot v_i + sumlimits_i = r + 1^n alpha_i cdot v_i right) &= sumlimits_i = 1^r beta_i cdot v_i + sumlimits_i = r + 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^r 1 cdot left( alpha_i cdot v_i right) + sumlimits_i = r + 1^n 1 cdot left( alpha_i cdot v_i right) &= sumlimits_i = 1^r beta_i cdot v_i + sumlimits_i = r + 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^r left( 1 alpha_i right) cdot v_i + sumlimits_i = r + 1^n left( 1 alpha_i right) cdot v_i &= sumlimits_i = 1^r beta_i cdot v_i + sumlimits_i = r + 1^m beta_i cdot w_i \
therefore sumlimits_i = 1^r alpha_i cdot v_i + sumlimits_i = r + 1^n alpha_i cdot v_i &= sumlimits_i = 1^r beta_i cdot v_i + sumlimits_i = r + 1^m beta_i cdot w_i \
endalign*
Adding the additive inverses of each of the vectors on the right hand side of the equation to both the sides and using axioms of vector space multiple times, we get
$$sumlimits_i = 1^r left( alpha_i - beta_i right) cdot v_i + sumlimits_i = r + 1^n alpha_i cdot v_i + sumlimits_i = r + 1^m left( - beta_i right) cdot w_i = textbf0$$
Since $S_1 cup S_2$ is linearly independent, $forall i in leftlbrace 1, 2, cdots, r rightrbrace, alpha_i - beta_i = 0 Rightarrow alpha_i = beta_i$. Also, $forall i in leftlbrace r + 1, r + 2, n cdots rightrbrace, alpha_i = 0$ and $forall i in leftlbrace r + 1, r + 2, cdots, m rightrbrace, beta_i = 0$. This tells us that $v = sumlimits_i = 1^r alpha_i cdot v_i$ and $w = 1 cdot v = sumlimits_i = 1^r alpha_i cdot v_i$. Hence, $1 cdot v = v$.

Note that we can prove $- left( a cdot v right) = left( -a right) cdot v$ as follows

beginalign
left( -a right) cdot v + a cdot v = left( -a + a right) cdot v = 0 cdot v = 0
endalign

Let $Bbb F=Bbb R$, $V=Bbb Z$ and for $alphainBbb F$, $vinBbb Z$, let $alpha v=0$. Then $v$ is an abelian group and $cdot$ is an action of the ring-without-$1$ (sometimes called rng) $Bbb F$ on $V$. Hence this is almost a vector space - only the fact that $1$ acts as identity is missing.

The $V$ does not have a basis. Indeed, any linear combination $alpha_1v_1+alpha_2v_2+ldots +a_nv_n$ will always be $=0$. So the very first sentence in your argument is invalid.