Bayes theorem: Can some one explain in plain words how argmax is used here?

Question

According to Bayes theorem

$$p(y |x) = \frac{p(x |y) p(y)}{p(x)}$$

So, to find the maximal value for $p(y |x)$, we want to compute:

\begin{align} &\arg\max_{y\in Y} p(y|x) \\ =&\arg\max_{y\in Y} \frac{p(x |y) p(y)}{p(x)}\\ =&\arg\max_{y\in Y}p(x |y) p(y) \end{align}

So, why we are ignoring $p(x)$ on the last line here?? Can some one explain in plain words?

Answer 1

The statement is not saying that $$\frac{p(x|y)p(y)}{p(x)} \underbrace{=}_{\text{Not true!}} p(x|y)p(y). $$ However, the $y$ that maximizes the former term is the same $y$ that maximizes the numerator, since the denominator does not depend on $y$.

Answer 2

Because the term $p(x)$ is same for $\forall y \in Y$.