Why is $P$ singular

Question

This is from Shafarevich's 'Basic Algebraic Geometry 1':

Let $P=(\alpha,\beta)\in X$, and suppose that the equation of $X$ is written in the form $f(x,y)=a(x-\alpha)+b(x-\beta)+g$, where $g$ is a polynomial expanded in the powers of $x-\alpha$ and $y-\beta$ with degree $\geq 2$.
The equation of a line $L$ through $P$ is in the form $x=\alpha + \lambda t$, $y=\beta + \mu t$.
$t$ local parameter on $L$ at $P$.
The restriction of $f$ to $L$: $f(\alpha+\lambda t, \beta + \mu t) = (a\lambda + b\mu)t+t^2 \varphi(t)$.
From this we see that if $P$ is singular, that is, if $a=b=0$, then every line through $P$ has intersection multiplicity $>1$.

Now my question, why do we know that for $P$ singular $a$ and $b$ must be zero?
And what is the definition of the intersection multiplicity? In my lecture we did not define it.

I would be really happy if someone could explain me the above.
Thanks a lot and all the best.

Answer 1

(I assume $X$ is meant to be an affine plane curve.)

To say that $X$ is singular at $P$ means that the dimension of the cotangent space at that point is bigger than it should be: that is,

$$\operatorname{dim }_k \mathfrak m/ \mathfrak m^2 > 1$$

where $\mathfrak m = \mathfrak m_{X,P}$ is the maximal ideal of the local ring of $X$ at $P$.

Now $X$ is the vanishing locus of $f$: working through the definitions, you find that $\mathfrak m / \mathfrak m^2$ is exactly the quotient of the cotangent space of the plane at $P$ by the element $df$:

$$ \mathfrak m/ \mathfrak m^2 = \left( \mathfrak M \,/ \,\mathfrak M^2 \right) / (df)$$

where $\mathfrak M$ is the maximal ideal of the local ring of $\mathbf A^2$ at $P$, and $df$ is the image of $f$ in $\mathfrak M \, /\, \mathfrak M^2$.

So $X$ is singular at $P$ if and only if $df=0$ in $\mathfrak M \, /\, \mathfrak M^2$, that is, if and only if $f \in \mathfrak M^2$, which happens if and only if both $a$ and $b$ are zero.

About intersection multiplicity: see Chapter IV of Volume 1 of Shafarevich. The definition in this case is pretty simple. Suppose two curves $C$ and $D$ defined by equations $f$ and $g$ meet properly at the point $P$. Let $\mathcal O$ be the local ring of the plane at $P$, and look at the vector space $\mathcal O/(f,g)$. Then the intersection multiplicity of the two curves at $P$ is defined as

$$(C \cdot D)_P = \operatorname{dim}_k \mathcal O / (f,g).$$

You can check that this number really is $>1$ in your case, using the argument you quoted in the question.