If $R$ is a commutative ring with unit element, then $R$ has a maximal ideal

Question

There is a theorem we stated during lectures and referred to it as "Kronecker's theorem". The theorem claims what is written in the title above, but proof to the theorem was messy, and I cannot find it anywhere. The proof looked something like this, and is based on Zorn's lemma: Let $(I_t)_{t \in T}$ is set of all proper ideals in $R$. It is inductively ordered set relative to inclusion (I do not know why this is true). For any two $t_1 \ne t_2$ is $I_{t_1} \subset I_{t_2}$ or $I_{t_2} \subset I_{t_1}$ $(1)$. Let $I = \bigcup\limits_{t \in T} I_t$. Let us prove $I$ is ideal in $R$. Let $x, y \in I$, then $x \in I_{t_1}$ and $y \in I_{t_2}$; so because of $(1)$ we get, without loss of generality, that $x, y \in I_{t_1}$ (for instance). But $I_{t_1}$ is ideal in $R$, so $x - y, xr \in I_{t_1}$, for any $r \in R$ and hence $I_{t_1} \subset I$, we obtain that $I$ is ideal in $R$. It is obvious that $I \ne R$, because $(I_t)_{t \in T}$ is family of proper ideals, so $1 \notin I_{t}, \forall t \in T$, which implies that $1 \notin I = \bigcup\limits_{t \in T} I_t$. Now, by applying Zorn's lemma on $I$ we get the claim of the theorem. I am confused about the claim that the family of the proper ideals is inductively ordered. Any help is appreciated.

Answer 1

You simply misunderstood what Zorn's lemma asserts: it says that if in a partially ordered set, every totally ordered subset has an upperbound, then the partially ordered set has maximal elements.

So here , what is checked if that any totally ordered family of ideals has an upper bound (which is, of course their union). This does not mean any family of ideals is totally ordered.