fix phi function link

jakobkogler · jakobkogler · commit 770c64fad558 · 2021-11-28T16:50:37.000+01:00
diff --git a/src/algebra/phi-function.md b/src/algebra/phi-function.md
@@ -23,7 +23,7 @@ The following properties of Euler totient function are sufficient to calculate i
 
 <ul><li>If $a$ and $b$ are relatively prime, then:
     $$\phi(a b) = \phi(a) \cdot \phi(b).$$
-    This relation is not trivial to see. It follows from the [Chinese remainder theorem](algebra/chinese-remainder-theorem.html). The Chinese remainder theorem guarantees, that for each $0 \le x < a$ and each $0 \le y < b$, there exists a unique $0 \le z < a b$ with $z \equiv x \pmod{a}$ and $z \equiv y \pmod{b}$. It's not hard to show that $z$ is coprime to $a b$ if and only if $x$ is coprime to $a$ and $y$ is coprime to $b$. Therefore the amount of integers coprime to $a b$ is equal to product of the amounts of $a$ and $b$.
+    This relation is not trivial to see. It follows from the <a href="algebra/chinese-remainder-theorem.html">Chinese remainder theorem</a>. The Chinese remainder theorem guarantees, that for each $0 \le x < a$ and each $0 \le y < b$, there exists a unique $0 \le z < a b$ with $z \equiv x \pmod{a}$ and $z \equiv y \pmod{b}$. It's not hard to show that $z$ is coprime to $a b$ if and only if $x$ is coprime to $a$ and $y$ is coprime to $b$. Therefore the amount of integers coprime to $a b$ is equal to product of the amounts of $a$ and $b$.
 </li>
 
 <li>
