Every polynomial has a root in some extension field - why is irreducibility needed in construction?

Question

Context: given a field $k$ and a polynomial $f(x)$ in $k[x]$, there exists an extension field $E$ of $k$ in which $f$ has a root. (note: this is all from Lang, Algebra, page 230, Algebraic Extensions)

Basic construction: we can ignore the fact that extension may not contain $k$ but rather isomorphic copy of $k$. Also, we need not consider $f(x)$ but only its irreducible factor $p(x)$, obviously. Construction goes using canonical homomorphism $$\sigma: k[x]\rightarrow k[x]/(p(x))$$ restricted on $k$, so the kernel is 0, setting $\xi = \sigma(x)$ (here $x$ is a polynomial, not indeterminate) and then showing that $\xi$ is a root of $\sigma(p(x))$. This is all clear to me.

Now the question: why is irreducibility even important here? Shouldn't the same construction pass just by using general $f(x)$? Sure, $k[x]/(p(x))$ is not a field anymore, but it seems to me that it doesn't matter, kernel of homomorphism is still 0.

Answer 1

You want to show that an extension field exists, hence the thing you exhibit better be a field. Exhibiting an extension ring (in general, even with zero divisors) does not help for the intended goal.

Answer 2

Have a look at what you wrote here (emphasis mine):

there exists an extension field $E$ of $k$ in which $f$ has a root.

Then compare that to what you wrote here (emphasis mine):

Sure, $k[x]/(p(x))$ is not a field anymore

That's exactly the point. We want to show that a certain kind of field exists. So we have to make sure our construction yields a field.